Finding Standard Deviation from MGF
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0
down vote
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I a working on a problem and am a little bit confused at how to approach solving it.
The problem: Given the MGF F(t) = $1over(1-2500t)^4$
Calculate the SD.
Do we need to do some substitution with like $mathscr N$ ~ (0,1) with sigma and mu? Or find the Var and square root it?
More broadly, how do we find the Standard Deviation of a Moment Generating Function?
probability standard-deviation moment-generating-functions moment-problem
asked Nov 17 at 20:45
Ethan
9212
add a comment |
up vote
0
down vote
favorite
I a working on a problem and am a little bit confused at how to approach solving it.
The problem: Given the MGF F(t) = $1over(1-2500t)^4$
Calculate the SD.
Do we need to do some substitution with like $mathscr N$ ~ (0,1) with sigma and mu? Or find the Var and square root it?
More broadly, how do we find the Standard Deviation of a Moment Generating Function?
probability standard-deviation moment-generating-functions moment-problem
asked Nov 17 at 20:45
Ethan
9212
SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I a working on a problem and am a little bit confused at how to approach solving it.
The problem: Given the MGF F(t) = $1over(1-2500t)^4$
Calculate the SD.
Do we need to do some substitution with like $mathscr N$ ~ (0,1) with sigma and mu? Or find the Var and square root it?
More broadly, how do we find the Standard Deviation of a Moment Generating Function?
probability standard-deviation moment-generating-functions moment-problem
asked Nov 17 at 20:45
Ethan
9212
I a working on a problem and am a little bit confused at how to approach solving it.
The problem: Given the MGF F(t) = $1over(1-2500t)^4$
Calculate the SD.
Do we need to do some substitution with like $mathscr N$ ~ (0,1) with sigma and mu? Or find the Var and square root it?
More broadly, how do we find the Standard Deviation of a Moment Generating Function?
probability standard-deviation moment-generating-functions moment-problem
probability standard-deviation moment-generating-functions moment-problem
asked Nov 17 at 20:45
Ethan
9212
asked Nov 17 at 20:45
Ethan
9212
edited Nov 17 at 21:24
asked Nov 17 at 20:45
Ethan
9212
asked Nov 17 at 20:45
Ethan
9212
asked Nov 17 at 20:45
Ethan
9212
9212
SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25
add a comment |
SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25
SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25
SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25
add a comment |
2 Answers
2
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oldest
votes
up vote
2
down vote
accepted
You don't find SD of a MGF ... you find SD of a random variable, cmon.
Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$
If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$
Now, SD = $sqrt{Var(X)} = sqrt{E[X^2] - E[X]^2}$.
Can you take it from there?
answered Nov 17 at 21:23
Makina
1,016113
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
add a comment |
up vote
0
down vote
Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $alpha=4$ and scale parameter $beta=2500$. The variance follows directly from the variance formula $alphabeta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $kappa(t)=log F(t)=-4log(1-2500t)$ and realize that the variance is the 2nd derivative of $kappa(t)$ evaluated at zero.
answered Nov 17 at 21:42
Knightgu
414
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You don't find SD of a MGF ... you find SD of a random variable, cmon.
Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$
If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$
Now, SD = $sqrt{Var(X)} = sqrt{E[X^2] - E[X]^2}$.
Can you take it from there?
answered Nov 17 at 21:23
Makina
1,016113
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
add a comment |
up vote
2
down vote
accepted
You don't find SD of a MGF ... you find SD of a random variable, cmon.
Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$
If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$
Now, SD = $sqrt{Var(X)} = sqrt{E[X^2] - E[X]^2}$.
Can you take it from there?
answered Nov 17 at 21:23
Makina
1,016113
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You don't find SD of a MGF ... you find SD of a random variable, cmon.
Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$
If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$
Now, SD = $sqrt{Var(X)} = sqrt{E[X^2] - E[X]^2}$.
Can you take it from there?
answered Nov 17 at 21:23
Makina
1,016113
You don't find SD of a MGF ... you find SD of a random variable, cmon.
Let $X$ be your random variable, then MGF of $X$ is basically $M_X(t) = E[e^{Xt}]$
If you check your textbook, you will find a nice property for the MGF, namely the n-th derivative of M evaluated at zero is equal to $E[X^n]$
Now, SD = $sqrt{Var(X)} = sqrt{E[X^2] - E[X]^2}$.
Can you take it from there?
answered Nov 17 at 21:23
Makina
1,016113
answered Nov 17 at 21:23
Makina
1,016113
answered Nov 17 at 21:23
Makina
1,016113
answered Nov 17 at 21:23
Makina
1,016113
1,016113
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
add a comment |
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
Thank you that makes sense.
– Ethan
Nov 17 at 21:25
add a comment |
up vote
0
down vote
Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $alpha=4$ and scale parameter $beta=2500$. The variance follows directly from the variance formula $alphabeta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $kappa(t)=log F(t)=-4log(1-2500t)$ and realize that the variance is the 2nd derivative of $kappa(t)$ evaluated at zero.
answered Nov 17 at 21:42
Knightgu
414
add a comment |
up vote
0
down vote
Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $alpha=4$ and scale parameter $beta=2500$. The variance follows directly from the variance formula $alphabeta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $kappa(t)=log F(t)=-4log(1-2500t)$ and realize that the variance is the 2nd derivative of $kappa(t)$ evaluated at zero.
answered Nov 17 at 21:42
Knightgu
414
add a comment |
up vote
0
down vote
up vote
0
down vote
Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $alpha=4$ and scale parameter $beta=2500$. The variance follows directly from the variance formula $alphabeta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $kappa(t)=log F(t)=-4log(1-2500t)$ and realize that the variance is the 2nd derivative of $kappa(t)$ evaluated at zero.
answered Nov 17 at 21:42
Knightgu
414
Another way is to recognize that the MGF belongs to the Gamma distribution with shape parameter $alpha=4$ and scale parameter $beta=2500$. The variance follows directly from the variance formula $alphabeta^2$ of the Gamma distribution. A more direct approach is to consider the cumulant generating function $kappa(t)=log F(t)=-4log(1-2500t)$ and realize that the variance is the 2nd derivative of $kappa(t)$ evaluated at zero.
answered Nov 17 at 21:42
Knightgu
414
answered Nov 17 at 21:42
Knightgu
414
answered Nov 17 at 21:42
Knightgu
414
answered Nov 17 at 21:42
Knightgu
414
414
add a comment |
add a comment |
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SD of the distribution, not of the MGF. Can you find moments like expectation and variance of the distribution from the MGF?
– StubbornAtom
Nov 17 at 21:25