How to prove $P(Acup B) = P(A) + P(B) - P(Acap B)$












2












$begingroup$


Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):



How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)



I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?



And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you should porably explain your notations.
    $endgroup$
    – J.F
    Jan 18 at 20:33










  • $begingroup$
    @G.F Iedited the question. Thanks.
    $endgroup$
    – Laura
    Jan 18 at 20:36






  • 1




    $begingroup$
    I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
    $endgroup$
    – WaveX
    Jan 18 at 20:37






  • 1




    $begingroup$
    It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
    $endgroup$
    – JMoravitz
    Jan 18 at 20:37












  • $begingroup$
    @JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
    $endgroup$
    – Laura
    Jan 18 at 20:41


















2












$begingroup$


Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):



How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)



I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?



And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you should porably explain your notations.
    $endgroup$
    – J.F
    Jan 18 at 20:33










  • $begingroup$
    @G.F Iedited the question. Thanks.
    $endgroup$
    – Laura
    Jan 18 at 20:36






  • 1




    $begingroup$
    I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
    $endgroup$
    – WaveX
    Jan 18 at 20:37






  • 1




    $begingroup$
    It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
    $endgroup$
    – JMoravitz
    Jan 18 at 20:37












  • $begingroup$
    @JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
    $endgroup$
    – Laura
    Jan 18 at 20:41
















2












2








2


2



$begingroup$


Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):



How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)



I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?



And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?






probability







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ any random variables. $A$ and $B$ are two events $in Omega$ (the sample space):



How can I prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ using the equation: $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],; A, B in Omega$ (my sample space)



I believe to prove this $E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)]$
I will have to test for $X>Y$, $X<Y$ and $X=Y$ right?



And to use this equation to prove $P(A cup B) = P(A) + P(B) - P(A cap B)$ I will have to use the idea of $A subset B, ; B subset A$ and $A = B,$ right?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 22:05









Andrés E. Caicedo

66.1k8160252




66.1k8160252










asked Jan 18 at 20:31









LauraLaura

3758




3758








  • 1




    $begingroup$
    you should porably explain your notations.
    $endgroup$
    – J.F
    Jan 18 at 20:33










  • $begingroup$
    @G.F Iedited the question. Thanks.
    $endgroup$
    – Laura
    Jan 18 at 20:36






  • 1




    $begingroup$
    I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
    $endgroup$
    – WaveX
    Jan 18 at 20:37






  • 1




    $begingroup$
    It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
    $endgroup$
    – JMoravitz
    Jan 18 at 20:37












  • $begingroup$
    @JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
    $endgroup$
    – Laura
    Jan 18 at 20:41
















  • 1




    $begingroup$
    you should porably explain your notations.
    $endgroup$
    – J.F
    Jan 18 at 20:33










  • $begingroup$
    @G.F Iedited the question. Thanks.
    $endgroup$
    – Laura
    Jan 18 at 20:36






  • 1




    $begingroup$
    I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
    $endgroup$
    – WaveX
    Jan 18 at 20:37






  • 1




    $begingroup$
    It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
    $endgroup$
    – JMoravitz
    Jan 18 at 20:37












  • $begingroup$
    @JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
    $endgroup$
    – Laura
    Jan 18 at 20:41










1




1




$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33




$begingroup$
you should porably explain your notations.
$endgroup$
– J.F
Jan 18 at 20:33












$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36




$begingroup$
@G.F Iedited the question. Thanks.
$endgroup$
– Laura
Jan 18 at 20:36




1




1




$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37




$begingroup$
I think you mean to write $$P(Acup B) = P(A) + P(B) - P(Acap B)$$ but you have it miswritten in your problem
$endgroup$
– WaveX
Jan 18 at 20:37




1




1




$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37






$begingroup$
It helps to note that $Acup B$ can be written as a disjoint union as $Acup B = (Asetminus B)cup (Acap B)cup (Bsetminus A)$ and similarly that $A = (Asetminus B)cup (Acap B)$, etc...
$endgroup$
– JMoravitz
Jan 18 at 20:37














$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41






$begingroup$
@JMoravitz but how can I use the idea from $E[max(X,Y)]=E[X]+E[Y]−E[min(X,Y)] $ proof?
$endgroup$
– Laura
Jan 18 at 20:41












2 Answers
2






active

oldest

votes


















6












$begingroup$

If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks @W-t-P! But how can I understand this concept of Indicator variable?
    $endgroup$
    – Laura
    Jan 19 at 1:20






  • 1




    $begingroup$
    @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
    $endgroup$
    – WaveX
    Jan 19 at 3:41





















1












$begingroup$

If you need to prove that
$$
E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
$$
simply note that the relation is valid even before you take expectations:
$$
max(X,Y) = X + Y - min(X,Y),
$$
which is the same as
$$
X + Y = min(X,Y) + max(X,Y),
$$
which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks @W-t-P! But how can I understand this concept of Indicator variable?
      $endgroup$
      – Laura
      Jan 19 at 1:20






    • 1




      $begingroup$
      @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
      $endgroup$
      – WaveX
      Jan 19 at 3:41


















    6












    $begingroup$

    If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks @W-t-P! But how can I understand this concept of Indicator variable?
      $endgroup$
      – Laura
      Jan 19 at 1:20






    • 1




      $begingroup$
      @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
      $endgroup$
      – WaveX
      Jan 19 at 3:41
















    6












    6








    6





    $begingroup$

    If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.






    share|cite|improve this answer









    $endgroup$



    If $X$ is the indicator variable of $A$ and $Y$ is the indicator variable of $B$, then $E(X)=P(A)$ and $E(Y)=P(B)$. Furthermore, $max{X,Y}$ is the indicator variable of $Acup B$, and $min{A,B}$ of $Acap B$; thus, $E(max{X,Y})=P(Acup B)$ and $E(min{X,Y})=P(Acap B)$. Therefore, with this choice of $X$ and $Y$, the relation $E(max{X,Y})=E(X)+E(Y)-E(min{X,Y})$ can be rewritten as $P(Acup B)=P(A)+P(B)-P(Acap B)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 at 20:44









    W-t-PW-t-P

    1,954612




    1,954612












    • $begingroup$
      thanks @W-t-P! But how can I understand this concept of Indicator variable?
      $endgroup$
      – Laura
      Jan 19 at 1:20






    • 1




      $begingroup$
      @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
      $endgroup$
      – WaveX
      Jan 19 at 3:41




















    • $begingroup$
      thanks @W-t-P! But how can I understand this concept of Indicator variable?
      $endgroup$
      – Laura
      Jan 19 at 1:20






    • 1




      $begingroup$
      @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
      $endgroup$
      – WaveX
      Jan 19 at 3:41


















    $begingroup$
    thanks @W-t-P! But how can I understand this concept of Indicator variable?
    $endgroup$
    – Laura
    Jan 19 at 1:20




    $begingroup$
    thanks @W-t-P! But how can I understand this concept of Indicator variable?
    $endgroup$
    – Laura
    Jan 19 at 1:20




    1




    1




    $begingroup$
    @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
    $endgroup$
    – WaveX
    Jan 19 at 3:41






    $begingroup$
    @Laura an Indicator Variable is a binary variable: it only takes the value $0$ or $1$. For example, $X$ can be thought to be an indicator for whether or not event $A$ had happened; if it did, then $X$ will take the value $1$, and if event $A$ does not happen, it is $0$
    $endgroup$
    – WaveX
    Jan 19 at 3:41













    1












    $begingroup$

    If you need to prove that
    $$
    E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
    $$
    simply note that the relation is valid even before you take expectations:
    $$
    max(X,Y) = X + Y - min(X,Y),
    $$
    which is the same as
    $$
    X + Y = min(X,Y) + max(X,Y),
    $$
    which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you need to prove that
      $$
      E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
      $$
      simply note that the relation is valid even before you take expectations:
      $$
      max(X,Y) = X + Y - min(X,Y),
      $$
      which is the same as
      $$
      X + Y = min(X,Y) + max(X,Y),
      $$
      which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you need to prove that
        $$
        E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
        $$
        simply note that the relation is valid even before you take expectations:
        $$
        max(X,Y) = X + Y - min(X,Y),
        $$
        which is the same as
        $$
        X + Y = min(X,Y) + max(X,Y),
        $$
        which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).






        share|cite|improve this answer









        $endgroup$



        If you need to prove that
        $$
        E[max(X,Y)] = E[X] + E[Y] - E[min(X,Y)],
        $$
        simply note that the relation is valid even before you take expectations:
        $$
        max(X,Y) = X + Y - min(X,Y),
        $$
        which is the same as
        $$
        X + Y = min(X,Y) + max(X,Y),
        $$
        which is true since you can find the sum of two numbers by adding the smaller number (the min) to the larger number (the max).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 22:50









        grand_chatgrand_chat

        20.5k11327




        20.5k11327






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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