independent, identically distributed (IID) random variables [closed]












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I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.



Would someone explain in simple terms what IID random variables are and give me an example?










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closed as unclear what you're asking by Did, Lord Shark the Unknown, max_zorn, Cesareo, metamorphy Jan 12 at 8:08


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















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    say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
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    – Eleven-Eleven
    Aug 13 '13 at 21:41
















12












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I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.



Would someone explain in simple terms what IID random variables are and give me an example?










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closed as unclear what you're asking by Did, Lord Shark the Unknown, max_zorn, Cesareo, metamorphy Jan 12 at 8:08


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















  • $begingroup$
    say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
    $endgroup$
    – Eleven-Eleven
    Aug 13 '13 at 21:41














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I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.



Would someone explain in simple terms what IID random variables are and give me an example?










share|cite|improve this question











$endgroup$




I am having trouble understanding IID random variables. I've tried reading http://scipp.ucsc.edu/~haber/ph116C/iid.pdf, http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture32.pdf, and http://www-inst.eecs.berkeley.edu/%7Ecs70/sp13/notes/n17.sp13.pdf but I don't get it.



Would someone explain in simple terms what IID random variables are and give me an example?







probability probability-theory random-variables independence






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edited Nov 12 '15 at 9:25









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asked Aug 13 '13 at 21:35









Frank EppsFrank Epps

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closed as unclear what you're asking by Did, Lord Shark the Unknown, max_zorn, Cesareo, metamorphy Jan 12 at 8:08


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Did, Lord Shark the Unknown, max_zorn, Cesareo, metamorphy Jan 12 at 8:08


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • $begingroup$
    say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
    $endgroup$
    – Eleven-Eleven
    Aug 13 '13 at 21:41


















  • $begingroup$
    say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
    $endgroup$
    – Eleven-Eleven
    Aug 13 '13 at 21:41
















$begingroup$
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
$endgroup$
– Eleven-Eleven
Aug 13 '13 at 21:41




$begingroup$
say you are sampling from a known distribution with replacement. Every time you, say, draw a sample, this is a random variable. Drawn samples are independent of each other, and the distribution never changes. Thus, IID random variables
$endgroup$
– Eleven-Eleven
Aug 13 '13 at 21:41










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Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.



If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.



They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.



They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).



Does that help?






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    An example of an identical dependent situation would help me :).
    $endgroup$
    – jiggunjer
    Sep 5 '16 at 5:46






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    @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
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    – Steven Stadnicki
    Nov 6 '16 at 2:41





















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"Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$



For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 cap X_2, X_3) = P(X_1 cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = frac{1}{6} cdot frac{1}{6}$, so are other combinations.



However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$ or $frac{2}{6} neq frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=frac{1}{3}$.



"Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc.



"Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.



"Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0.



"Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.






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    .So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer






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      Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
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      – Shailesh
      Mar 11 '16 at 14:46




















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    3 Answers
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    Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.



    If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.



    They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.



    They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).



    Does that help?






    share|cite|improve this answer











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    • $begingroup$
      An example of an identical dependent situation would help me :).
      $endgroup$
      – jiggunjer
      Sep 5 '16 at 5:46






    • 1




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      @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
      $endgroup$
      – Steven Stadnicki
      Nov 6 '16 at 2:41


















    27












    $begingroup$

    Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.



    If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.



    They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.



    They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).



    Does that help?






    share|cite|improve this answer











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    • $begingroup$
      An example of an identical dependent situation would help me :).
      $endgroup$
      – jiggunjer
      Sep 5 '16 at 5:46






    • 1




      $begingroup$
      @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
      $endgroup$
      – Steven Stadnicki
      Nov 6 '16 at 2:41
















    27












    27








    27





    $begingroup$

    Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.



    If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.



    They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.



    They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).



    Does that help?






    share|cite|improve this answer











    $endgroup$



    Im sure you know that iid means independent, identically distributed. I think the most prominent example is a coin toss repeated several times.



    If $X_1, X_2, dots$ designate the result of the 1st, 2nd, and so on toss (where $X_i = 1$ means that in the i-th toss you have got head and $X_i = 0$ tail), you have that $X_1,X_2,dots $ are iid.



    They are independent since every time you flip a coin, the previous result doesn't influence your current result. Edit: there is a mathematical definition of independence, but I don't think it is necessary at the moment.



    They are identically distributed, since every time you flip a coin, the chances of getting head (or tail) are identical, no matter if its the 1st or the 100th toss (probability distribution is identical over time). If the coin is "fair" the chances are 0,5 for each event (getting head or tail).



    Does that help?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 13 '13 at 23:41

























    answered Aug 13 '13 at 21:41









    Dima McGreenDima McGreen

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      An example of an identical dependent situation would help me :).
      $endgroup$
      – jiggunjer
      Sep 5 '16 at 5:46






    • 1




      $begingroup$
      @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
      $endgroup$
      – Steven Stadnicki
      Nov 6 '16 at 2:41




















    • $begingroup$
      An example of an identical dependent situation would help me :).
      $endgroup$
      – jiggunjer
      Sep 5 '16 at 5:46






    • 1




      $begingroup$
      @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
      $endgroup$
      – Steven Stadnicki
      Nov 6 '16 at 2:41


















    $begingroup$
    An example of an identical dependent situation would help me :).
    $endgroup$
    – jiggunjer
    Sep 5 '16 at 5:46




    $begingroup$
    An example of an identical dependent situation would help me :).
    $endgroup$
    – jiggunjer
    Sep 5 '16 at 5:46




    1




    1




    $begingroup$
    @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
    $endgroup$
    – Steven Stadnicki
    Nov 6 '16 at 2:41






    $begingroup$
    @jiggunjer How about this: Let $X_1$ be the result of flipping a coin, then let $X_{i+1}=X_i$ for all $i$. The $X_i$ are identically distributed - each variable has .5 chance of being 0 and .5 chance of being 1 - but they're as correlated as can possibly be.
    $endgroup$
    – Steven Stadnicki
    Nov 6 '16 at 2:41













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    "Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$



    For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 cap X_2, X_3) = P(X_1 cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = frac{1}{6} cdot frac{1}{6}$, so are other combinations.



    However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$ or $frac{2}{6} neq frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=frac{1}{3}$.



    "Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc.



    "Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.



    "Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0.



    "Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.






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      5












      $begingroup$

      "Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$



      For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 cap X_2, X_3) = P(X_1 cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = frac{1}{6} cdot frac{1}{6}$, so are other combinations.



      However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$ or $frac{2}{6} neq frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=frac{1}{3}$.



      "Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc.



      "Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.



      "Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0.



      "Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.






      share|cite|improve this answer











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        5








        5





        $begingroup$

        "Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$



        For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 cap X_2, X_3) = P(X_1 cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = frac{1}{6} cdot frac{1}{6}$, so are other combinations.



        However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$ or $frac{2}{6} neq frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=frac{1}{3}$.



        "Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc.



        "Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.



        "Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0.



        "Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.






        share|cite|improve this answer











        $endgroup$



        "Independent" means for any $x_i in X$, $P(x_0, x_1,..., x_i) = prod_0^i P(x_i)$



        For example, toss 2 dice. Let $X_1$ be event of the first being {1, 2}, $X_2$ the first being {2, 3, 4} and $X_3$ the second being {6} and $X_4$ the second being {4, 6}. It's intuitive to conclude that $P(X_1 cap X_2, X_3) = P(X_1 cap X_2) P(X_3) = P(X_1=2) P(X_3)= P(X_2=2) P(X_3) = frac{1}{6} cdot frac{1}{6}$, so are other combinations.



        However, synonyms of "identical" include "alike" and "equal". That's, the probability of every variable should be equal, or identical. In the above example $P(X_1) neq P(X_2)$ or $frac{2}{6} neq frac{3}{6}$. To make $X_1$ and $X_2$ be indentical variables, we can let $X_2$ be event of the second die being {1, 6} or the first being {4, 2} or other similar posibilities you like. Then we get $P(X_1) = P(X_2)=frac{1}{3}$.



        "Independent but not identical" as shown in the above two instances, like any combinations of evens between two dice: $P(X_1)$ and $P(X_3)$ or $P(X_2)$ and $P(X_4)$ and etc.



        "Identical and independent", any event combinations between the two dice that have the same probability. For example $P(X_1=1)$ and $P(X_3=6)$ and etc.



        "Identical but not independent", any two events for one die. For instance, $P(X_1=1)$ and $P(X_2=3)$. It is not independent since if the die shows you 1 you can not see a 3 at the same time, and hence the probability of seeing 1 and 3 at the same time while rolling one die is 0.



        "Not identical and not independent", any two events for one die with unequal probability, for instance $P(X_1)$ and $P(X_2)$ or $P(X_3)$ and $P(X_4)$, or $P(X_2=4)$ and $P(X_1)$ and etc.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 11 at 14:06

























        answered Nov 6 '16 at 2:24









        lernerlerner

        314217




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            .So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
              $endgroup$
              – Shailesh
              Mar 11 '16 at 14:46


















            0












            $begingroup$

            .So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
              $endgroup$
              – Shailesh
              Mar 11 '16 at 14:46
















            0












            0








            0





            $begingroup$

            .So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer






            share|cite|improve this answer









            $endgroup$



            .So basically you will consider events where the outcome in one case will not depend on the outcome of the other cases .It is called identical because in every case u consider the possible outcomes will be same as the previous event .Some one has suggested yes tossing of coin is a good example .I will try to be a statistician here .You should go through few statistical distributions like normal ,gamma etc and then see additive property .There while proving the additive property u will consider independent events initially and then prove that the addition of all the independent variates also follows the respective distribution using the MGF of that particular distribution and then you will extend your property to see what if the variates are made similar .Hope you got your answer







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 11 '16 at 14:26









            Tejas SureshTejas Suresh

            111




            111












            • $begingroup$
              Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
              $endgroup$
              – Shailesh
              Mar 11 '16 at 14:46




















            • $begingroup$
              Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
              $endgroup$
              – Shailesh
              Mar 11 '16 at 14:46


















            $begingroup$
            Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
            $endgroup$
            – Shailesh
            Mar 11 '16 at 14:46






            $begingroup$
            Welcome to Math.SE. This question has a well-accepted answer. You have provided nothing new. And what you have provided does not exactly answer the question. Please refrain from opening up old questions which have good answers unless you have something significant to contribute. THere are many new questions begging for answers.
            $endgroup$
            – Shailesh
            Mar 11 '16 at 14:46





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