Moment generating function of i.i.d
0
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I was reading this pdf https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2003/lecture-notes/lec15.pdf
I have two questions
- I know about moment generating function of say X1, then how was the formula for the iid sequence X1...Xn derived?
- How did the author do
$$Pi_{i=1}^nEe^{t_iX_i}=Pi_{i=1}^ne^{frac{t_i^2}{2}}?$$
statistics random-variables moment-generating-functions
asked Dec 9 '18 at 16:25
q126yq126y
239212
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add a comment |
0
$begingroup$
I was reading this pdf https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2003/lecture-notes/lec15.pdf
I have two questions
- I know about moment generating function of say X1, then how was the formula for the iid sequence X1...Xn derived?
- How did the author do
$$Pi_{i=1}^nEe^{t_iX_i}=Pi_{i=1}^ne^{frac{t_i^2}{2}}?$$
statistics random-variables moment-generating-functions
asked Dec 9 '18 at 16:25
q126yq126y
239212
$endgroup$
add a comment |
0
0
0
$begingroup$
I was reading this pdf https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2003/lecture-notes/lec15.pdf
I have two questions
- I know about moment generating function of say X1, then how was the formula for the iid sequence X1...Xn derived?
- How did the author do
$$Pi_{i=1}^nEe^{t_iX_i}=Pi_{i=1}^ne^{frac{t_i^2}{2}}?$$
statistics random-variables moment-generating-functions
asked Dec 9 '18 at 16:25
q126yq126y
239212
$endgroup$
I was reading this pdf https://ocw.mit.edu/courses/mathematics/18-443-statistics-for-applications-fall-2003/lecture-notes/lec15.pdf
I have two questions
- I know about moment generating function of say X1, then how was the formula for the iid sequence X1...Xn derived?
- How did the author do
$$Pi_{i=1}^nEe^{t_iX_i}=Pi_{i=1}^ne^{frac{t_i^2}{2}}?$$
statistics random-variables moment-generating-functions
statistics random-variables moment-generating-functions
asked Dec 9 '18 at 16:25
q126yq126y
239212
asked Dec 9 '18 at 16:25
q126yq126y
239212
asked Dec 9 '18 at 16:25
q126yq126y
239212
asked Dec 9 '18 at 16:25
q126yq126y
239212
asked Dec 9 '18 at 16:25
q126yq126y
239212
239212
add a comment |
add a comment |
1 Answer
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- Let $X_i$ have cdf $F_X(x_i)$, so its mgf is $int_{Bbb R}exp t_iX_i dF_X(x_i)$. By independence, the mgf of $sum_i X_i$ is $int_{Bbb R^n}expsum_i t_iX_i prod_idF_X(x_i)=prod_iint_{Bbb R}exp t_iX_i dF_X(x_i)$. In other words, we just multiply the characteristic functions.
- The claimed result follows exactly if each $X_isim N(0,,1)$, as then $Eexp t_iX_i=exp t_i^2/2$. (For more general distributions of mean $0$ and variance $1$, the central limit theorem implies the result is approximately valid.) Indeed, the substitution $y=x-t$ gives $$int_{Bbb R}frac{1}{sqrt{2pi}}exp (tx-tfrac{x^2}{2})dx=int_{Bbb R}frac{1}{sqrt{2pi}}exptfrac{t^2-y^2}{2}dy=exptfrac{t^2}{2}.$$
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
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Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
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– q126y
Dec 10 '18 at 15:14
add a comment |
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$begingroup$
- Let $X_i$ have cdf $F_X(x_i)$, so its mgf is $int_{Bbb R}exp t_iX_i dF_X(x_i)$. By independence, the mgf of $sum_i X_i$ is $int_{Bbb R^n}expsum_i t_iX_i prod_idF_X(x_i)=prod_iint_{Bbb R}exp t_iX_i dF_X(x_i)$. In other words, we just multiply the characteristic functions.
- The claimed result follows exactly if each $X_isim N(0,,1)$, as then $Eexp t_iX_i=exp t_i^2/2$. (For more general distributions of mean $0$ and variance $1$, the central limit theorem implies the result is approximately valid.) Indeed, the substitution $y=x-t$ gives $$int_{Bbb R}frac{1}{sqrt{2pi}}exp (tx-tfrac{x^2}{2})dx=int_{Bbb R}frac{1}{sqrt{2pi}}exptfrac{t^2-y^2}{2}dy=exptfrac{t^2}{2}.$$
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
$endgroup$
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
add a comment |
1
$begingroup$
- Let $X_i$ have cdf $F_X(x_i)$, so its mgf is $int_{Bbb R}exp t_iX_i dF_X(x_i)$. By independence, the mgf of $sum_i X_i$ is $int_{Bbb R^n}expsum_i t_iX_i prod_idF_X(x_i)=prod_iint_{Bbb R}exp t_iX_i dF_X(x_i)$. In other words, we just multiply the characteristic functions.
- The claimed result follows exactly if each $X_isim N(0,,1)$, as then $Eexp t_iX_i=exp t_i^2/2$. (For more general distributions of mean $0$ and variance $1$, the central limit theorem implies the result is approximately valid.) Indeed, the substitution $y=x-t$ gives $$int_{Bbb R}frac{1}{sqrt{2pi}}exp (tx-tfrac{x^2}{2})dx=int_{Bbb R}frac{1}{sqrt{2pi}}exptfrac{t^2-y^2}{2}dy=exptfrac{t^2}{2}.$$
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
$endgroup$
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
add a comment |
1
1
1
$begingroup$
- Let $X_i$ have cdf $F_X(x_i)$, so its mgf is $int_{Bbb R}exp t_iX_i dF_X(x_i)$. By independence, the mgf of $sum_i X_i$ is $int_{Bbb R^n}expsum_i t_iX_i prod_idF_X(x_i)=prod_iint_{Bbb R}exp t_iX_i dF_X(x_i)$. In other words, we just multiply the characteristic functions.
- The claimed result follows exactly if each $X_isim N(0,,1)$, as then $Eexp t_iX_i=exp t_i^2/2$. (For more general distributions of mean $0$ and variance $1$, the central limit theorem implies the result is approximately valid.) Indeed, the substitution $y=x-t$ gives $$int_{Bbb R}frac{1}{sqrt{2pi}}exp (tx-tfrac{x^2}{2})dx=int_{Bbb R}frac{1}{sqrt{2pi}}exptfrac{t^2-y^2}{2}dy=exptfrac{t^2}{2}.$$
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
$endgroup$
- Let $X_i$ have cdf $F_X(x_i)$, so its mgf is $int_{Bbb R}exp t_iX_i dF_X(x_i)$. By independence, the mgf of $sum_i X_i$ is $int_{Bbb R^n}expsum_i t_iX_i prod_idF_X(x_i)=prod_iint_{Bbb R}exp t_iX_i dF_X(x_i)$. In other words, we just multiply the characteristic functions.
- The claimed result follows exactly if each $X_isim N(0,,1)$, as then $Eexp t_iX_i=exp t_i^2/2$. (For more general distributions of mean $0$ and variance $1$, the central limit theorem implies the result is approximately valid.) Indeed, the substitution $y=x-t$ gives $$int_{Bbb R}frac{1}{sqrt{2pi}}exp (tx-tfrac{x^2}{2})dx=int_{Bbb R}frac{1}{sqrt{2pi}}exptfrac{t^2-y^2}{2}dy=exptfrac{t^2}{2}.$$
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
answered Dec 9 '18 at 16:36
J.G.J.G.
27.1k22843
27.1k22843
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
add a comment |
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
$begingroup$
Thanks. Can you please take a look at this question I posted math.stackexchange.com/questions/3033702/…
$endgroup$
– q126y
Dec 10 '18 at 15:14
add a comment |
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