# Math Help - MGF normal distribution

1. ## MGF normal distribution

Hello everyone

I looking at wikipedia for normal distribution and doing some work on moment generating functions. Please could someone help me with full working from getting from

$e^\frac{\sigma^2{t^2}}{2}$

to formula that say

Thank you.

2. Originally Posted by jescavez
Hello everyone

I looking at wikipedia for normal distribution and doing some work on moment generating functions. Please could someone help me with full working from getting from

$e^\frac{\sigma^2{t^2}}{2}$

to formula that say

Thank you.
Look at the definition and properties of the MGF, you will find that:

$E(x^{2k})=(2k)!\; a_{2k}$

where $a_{2k}$ is the coefficient of $t^{2k}$ in the power series expansion of the MGF. We know the series expansion of $e^{\sigma^2t^2/2}$ so carry on from there.

RonL

3. Hi RonL

Thank you. Am I on the right track with this below then for normal dist MGF:

$1+tx\sigma^2+\frac{t^2\sigma^4x^2}{2!}+...$

I am checking the pages for MGF and normal distribution for this formula below:

$E(x^{2k})=(2k)!\; a_{2k}$

But I did not see it yet, where does this come from please?

In first point, am I doing power series expansion wrong? Because I do not know where
$2^n$ will come from for denominator in final equation or how to see that odd moments are 0.

Thank you again

4. Originally Posted by jescavez
Hi RonL

Thank you. Am I on the right track with this below then for normal dist MGF:

$1+tx\sigma^2+\frac{t^2\sigma^4x^2}{2!}+...$ Mr F says: Where have these x's come from? See main reply.

I am checking the pages for MGF and normal distribution for this formula below:

$E(x^{2k})=(2k)!\; a_{2k}$

But I did not see it yet, where does this come from please?

In first point, am I doing power series expansion wrong? Because I do not know where $2^n$ will come from for denominator in final equation or how to see that odd moments are 0.

Thank you again
The MGF you quoted in your original question is for X ~ N(0, $\sigma$): $m_X(t) = e^{\frac{\sigma^2 t^2}{2}}$. So I don't know where you got those x's from in your power series ....??

By definition, the (2k)th moment is given by $E(X^{2k}) = M^{(2k)}(0)$. That is, you have to differentiate the MGF 2k times and then substitute t = 0.

The power series is $m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....$

Note that the kth term of this power series is $\frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{2k} k!}$.

If you think about it, the answer you want is got by differentiating this term 2k times .....

5. Hello Mr F

Thank you for your help, I tell you where I get my x's!:
I put in x's because I thinking like this:

t is variable so i differentiate $e^{\frac{\sigma^2 t^2}{2}}$ with respect to t which I thought would give me $t\sigma^2$ each time so I would have:

$1 +t\sigma^2e^{\frac{\sigma^2 t^2}{2}}+t^2\sigma^4e^{\frac{\sigma^2 t^2}{2}}+t^3\sigma^6e^{\frac{\sigma^2 t^2}{2}}..$

which I equate against polynomial expression set up and differentiated at same time

$a + bx + cx^2 + dx^3$

I was confuse about this method because I trying to think what point I expand about which no make sense. Is it completely wrong to do this?

I was thinking that you need $e^{\frac{\sigma^2 t^2}{2}}$ to be $e^{\frac{\sigma^2 t^2x}{2}}$ so that you get ${\frac{\sigma^2 t^2}{2}}$ when you differentiate. Is this property of MGF that I don't understand maybe?
How does this prove that odd moments are not there?

Thank you again.

6. Originally Posted by mr fantastic

Note that the kth term of this power series is $\frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{2k} k!}$.
The kth term of this power series is $\frac{ \left( \frac{\sigma^2 t^2}{2} \right)^k}{k!} = \frac{\sigma^{2k} t^{2k}}{2^{k} k!}$.

7. Originally Posted by jescavez
Hello Mr F

Thank you for your help, I tell you where I get my x's!:
I put in x's because I thinking like this:

t is variable so i differentiate $e^{\frac{\sigma^2 t^2}{2}}$ with respect to t which I thought would give me $t\sigma^2$ each time so I would have:

$1 +t\sigma^2e^{\frac{\sigma^2 t^2}{2}}+t^2\sigma^4e^{\frac{\sigma^2 t^2}{2}}+t^3\sigma^6e^{\frac{\sigma^2 t^2}{2}}..$

which I equate against polynomial expression set up and differentiated at same time

$a + bx + cx^2 + dx^3$

I was confuse about this method because I trying to think what point I expand about which no make sense. Is it completely wrong to do this?

I was thinking that you need $e^{\frac{\sigma^2 t^2}{2}}$ to be $e^{\frac{\sigma^2 t^2x}{2}}$ so that you get ${\frac{\sigma^2 t^2}{2}}$ when you differentiate. Is this property of MGF that I don't understand maybe?
How does this prove that odd moments are not there?

Thank you again.
Sorry but you have the wrong idea. The MGF is function of t and so its power series expansion is a function of t. The result is given in my previous reply. The expansion is the standard expansion for an exponential function (it's expanded around the point t = 0). You might need to review this idea a bit more thoroughly.

The odd moments are zero because if you differentiate an odd number of times, all the terms in the power series expansion contain t. When you put t = 0 the result is therefore zero.

8. Thank you Mr F. Is there any place online that gives a detailed step by step explanation of MGF like this?

I see where $2k!$ come from now also, thanks.

9. Originally Posted by jescavez
Thank you Mr F. Is there any place online that gives a detailed step by step explanation of MGF like this?

I see where $2k!$ come from now also, thanks.
What steps do you still need in this question? They can be shown here ....

10. I guess the part where we say that

$
m_X(t) = e^{\frac{\sigma^2 t^2}{2}}
$

has expansion
$
m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....
$

I no understand that very well. I thought that MGF is proved by Power series expansion of $e^{tx}$ it seems here we just take exponent and leave as it is, why is it unaffacted?

I also do not know where this assumption come from:
$
E(X^{2k}) = M^{(2k)}(0)
$

11. Originally Posted by jescavez
I guess the part where we say that

$
m_X(t) = e^{\frac{\sigma^2 t^2}{2}}
$

has expansion
$
m_X(t) = 1 + \frac{\sigma^2 t^2}{2} + \frac{\left(\frac{\sigma^2 t^2}{2}\right)^2}{2!} + \, .... + \, \frac{\left(\frac{\sigma^2 t^2}{2}\right)^k}{k!} + \, ....
$

I no understand that very well. I thought that MGF is proved by Power series expansion of $e^{tx}$ it seems here we just take exponent and leave as it is, why is it unaffacted?

I also do not know where this assumption come from:
$
E(X^{2k}) = M^{(2k)}(0)
$
You know that the series expansion for e^w is $1 + w + \frac{w^2}{2!} + \frac{w^3}{3!} + \, ....... \, + \frac{w^k}{k!} + \, ......$.

So to get the series expansion of $m_X(t) = e^{\frac{\sigma^2 t^2}{2}}$ you can just make the substitution $w = \frac{\sigma^2 t^2}{2}$. Then you will get the result I stated earlier.

"I thought that MGF is proved by Power series expansion of $e^{tx}$ it seems here we just take exponent and leave as it is, why is it unaffacted?"

That is done as part of the proof that $E(X^{n}) = M^{(n)}(0)
$
.

Summary of general results:

1. The MGF of a random variable X is $m_X(t) = E \left( e^{tX} \right)$.

2. The nth moment of X , that is, $E\left(X^{n}\right)$ is given by $\frac{d^n m_X}{dt^n}$ evaluated at t = 0, that is, $E\left(X^n \right) = m_X^{(n)}(0)$.

In your problem, $m_X(t) = e^{\frac{\sigma^2 t^2}{2}}$ and n = 2k. It is convenient to get the (2k)th derivative of $m_X(t)$ by writing it as a power series. Alternatively, you could just differentiate $e^{\frac{\sigma^2 t^2}{2}}$ 2k times without using the expansion.

Do not confuse expressing the calculated MGF as a power series in order to easily do the differentiations with using a power series to prove that the nth moment of the MGF is given by the formula in 2. above.

I hope this clears things up for you.

12. Hi,

I'm working on what seems to be essentially the same problem. I'm stuck in one area.

I understand that the mgf can be written $\frac{t^{2k}}{(k!*2^k)} *\sigma^{2k}$

but I don't understand how to get to

the $t^{2k}$ -> $(2k)!$ is tripping me up.

Thanks!

13. Originally Posted by reflex
Hi,

I'm working on what seems to be essentially the same problem. I'm stuck in one area.

I understand that the mgf can be written $\frac{t^{2k}}{(k!*2^k)} *\sigma^{2k}$

but I don't understand how to get to

the $t^{2k}$ -> $(2k)!$ is tripping me up.

Thanks!
You need to differentiate with respect to t the mgf 2k times and then substitute t = 0. Please show your work.

14. In your problem, $m_X(t) = e^{\frac{\sigma^2 t^2}{2}}$ and n = 2k.

I understand now, I think. I just have one more question and I'm not sure if it makes sense. I understand it 'works out' if n = 2k, but how do we justify that?