show that E(Y ) = rE(X_1)












0












$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10


















0












$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10
















0












0








0





$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$




For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes




probability statistics stochastic-processes






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 0:16









bluemusebluemuse

976




976












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10




















  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10


















$begingroup$
This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
$endgroup$
– Makina
Jan 16 at 0:23






$begingroup$
This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
$endgroup$
– Makina
Jan 16 at 0:23














$begingroup$
@makina $r$ isn't random.
$endgroup$
– user608030
Jan 16 at 0:27




$begingroup$
@makina $r$ isn't random.
$endgroup$
– user608030
Jan 16 at 0:27












$begingroup$
hence the "similar"
$endgroup$
– Makina
Jan 16 at 0:41




$begingroup$
hence the "similar"
$endgroup$
– Makina
Jan 16 at 0:41












$begingroup$
@bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 6:10






$begingroup$
@bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 6:10












1 Answer
1






active

oldest

votes


















0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09
















0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09














0












0








0





$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$



If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 0:47









HenryHenry

101k482169




101k482169












  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09


















  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09
















$begingroup$
Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
$endgroup$
– bluemuse
Jan 16 at 0:52






$begingroup$
Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
$endgroup$
– bluemuse
Jan 16 at 0:52














$begingroup$
they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
$endgroup$
– k.dkhk
Jan 18 at 13:08




$begingroup$
they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
$endgroup$
– k.dkhk
Jan 18 at 13:08












$begingroup$
@k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
$endgroup$
– Henry
Jan 18 at 16:09




$begingroup$
@k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
$endgroup$
– Henry
Jan 18 at 16:09


















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