moment generating function for $S_N=X_1+cdots+X_N$ with $N$ dependent of $X_1$moment generating function for $S_N$Prove that $P_S_N(t) = P_N(P_X(t))$ for $S_N = X_1 + cdots + X_N$.Moment generating function for independent random variablesDistribution of $Z$ from Moment Generating FunctionMoment generating function and probabilityFormula for expectation $E[varphi(X_1,X_2,ldots,X_n)]$?Possible typo in my textbookIf $X_1,ldots,X_n$ are independent, does $mathbbPX_n>maxX_1,ldots,X_n-1=mathbbP{X_n>X_1)cdotsmathbbPX_n>X_n-1$ hold?Moment-generating function of $m$ independent variablesExample iid variables $X_i$ where $S=sum_j=1^NX_j$ but $M_S(t) neq P_N(M_X(t))$
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moment generating function for $S_N=X_1+cdots+X_N$ with $N$ dependent of $X_1$
moment generating function for $S_N$Prove that $P_S_N(t) = P_N(P_X(t))$ for $S_N = X_1 + cdots + X_N$.Moment generating function for independent random variablesDistribution of $Z$ from Moment Generating FunctionMoment generating function and probabilityFormula for expectation $E[varphi(X_1,X_2,ldots,X_n)]$?Possible typo in my textbookIf $X_1,ldots,X_n$ are independent, does $mathbbPX_n>maxX_1,ldots,X_n-1=mathbbP{X_n>X_1)cdotsmathbbPX_n>X_n-1$ hold?Moment-generating function of $m$ independent variablesExample iid variables $X_i$ where $S=sum_j=1^NX_j$ but $M_S(t) neq P_N(M_X(t))$
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Let $X_1, X_2, ldots$ independent and identically distributed discrete random variables with support contained in $mathbbN$ and let $N=X_1+1$. How can I calculate the moment generating function of
$$S_N=X_1+cdots+X_N?$$
Thanks for any help.
My attepmt: We have that
$$M_S_N(t)=E[e^rS_N]=E[e^r0]Pr(N=0)+sum_n=1^infty E[e^r(X_1+cdots+X_n)|N=n]Pr(N=n) $$
probability probability-theory moment-generating-functions
asked Mar 29 at 10:20
krenickkrenick
183
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, ldots$ independent and identically distributed discrete random variables with support contained in $mathbbN$ and let $N=X_1+1$. How can I calculate the moment generating function of
$$S_N=X_1+cdots+X_N?$$
Thanks for any help.
My attepmt: We have that
$$M_S_N(t)=E[e^rS_N]=E[e^r0]Pr(N=0)+sum_n=1^infty E[e^r(X_1+cdots+X_n)|N=n]Pr(N=n) $$
probability probability-theory moment-generating-functions
asked Mar 29 at 10:20
krenickkrenick
183
$endgroup$
$begingroup$
If you provide some context and share your attempts on solving the problem you are more likely to get responses.
$endgroup$
– PierreCarre
Mar 29 at 10:26
add a comment |
$begingroup$
Let $X_1, X_2, ldots$ independent and identically distributed discrete random variables with support contained in $mathbbN$ and let $N=X_1+1$. How can I calculate the moment generating function of
$$S_N=X_1+cdots+X_N?$$
Thanks for any help.
My attepmt: We have that
$$M_S_N(t)=E[e^rS_N]=E[e^r0]Pr(N=0)+sum_n=1^infty E[e^r(X_1+cdots+X_n)|N=n]Pr(N=n) $$
probability probability-theory moment-generating-functions
asked Mar 29 at 10:20
krenickkrenick
183
$endgroup$
Let $X_1, X_2, ldots$ independent and identically distributed discrete random variables with support contained in $mathbbN$ and let $N=X_1+1$. How can I calculate the moment generating function of
$$S_N=X_1+cdots+X_N?$$
Thanks for any help.
My attepmt: We have that
$$M_S_N(t)=E[e^rS_N]=E[e^r0]Pr(N=0)+sum_n=1^infty E[e^r(X_1+cdots+X_n)|N=n]Pr(N=n) $$
probability probability-theory moment-generating-functions
probability probability-theory moment-generating-functions
asked Mar 29 at 10:20
krenickkrenick
183
asked Mar 29 at 10:20
krenickkrenick
183
edited Mar 29 at 10:35
krenick
asked Mar 29 at 10:20
krenickkrenick
183
asked Mar 29 at 10:20
krenickkrenick
183
asked Mar 29 at 10:20
krenickkrenick
183
183
$begingroup$
If you provide some context and share your attempts on solving the problem you are more likely to get responses.
$endgroup$
– PierreCarre
Mar 29 at 10:26
add a comment |
$begingroup$
If you provide some context and share your attempts on solving the problem you are more likely to get responses.
$endgroup$
– PierreCarre
Mar 29 at 10:26
$begingroup$
If you provide some context and share your attempts on solving the problem you are more likely to get responses.
$endgroup$
– PierreCarre
Mar 29 at 10:26
$begingroup$
If you provide some context and share your attempts on solving the problem you are more likely to get responses.
$endgroup$
– PierreCarre
Mar 29 at 10:26
add a comment |
1 Answer
1
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$begingroup$
$Ee^tS_N=sum_n EI_X_1=ne^t(X_1+X_2+..+X_n+1)=sum_n M(t)^nP(X_1=n)e^tn=sum_n P(X_1=n)e^(t+log(M(t))n=M(t+log(M(t))$.
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$Ee^tS_N=sum_n EI_X_1=ne^t(X_1+X_2+..+X_n+1)=sum_n M(t)^nP(X_1=n)e^tn=sum_n P(X_1=n)e^(t+log(M(t))n=M(t+log(M(t))$.
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
add a comment |
$begingroup$
$Ee^tS_N=sum_n EI_X_1=ne^t(X_1+X_2+..+X_n+1)=sum_n M(t)^nP(X_1=n)e^tn=sum_n P(X_1=n)e^(t+log(M(t))n=M(t+log(M(t))$.
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
add a comment |
$begingroup$
$Ee^tS_N=sum_n EI_X_1=ne^t(X_1+X_2+..+X_n+1)=sum_n M(t)^nP(X_1=n)e^tn=sum_n P(X_1=n)e^(t+log(M(t))n=M(t+log(M(t))$.
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
$endgroup$
$Ee^tS_N=sum_n EI_X_1=ne^t(X_1+X_2+..+X_n+1)=sum_n M(t)^nP(X_1=n)e^tn=sum_n P(X_1=n)e^(t+log(M(t))n=M(t+log(M(t))$.
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
answered Mar 29 at 10:28
Kavi Rama MurthyKavi Rama Murthy
72.5k53170
72.5k53170
add a comment |
add a comment |
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If you provide some context and share your attempts on solving the problem you are more likely to get responses.
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– PierreCarre
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