# Thread: Handling a moment generating function

1. ## Handling a moment generating function

In my problem, I have the moment-generating function on a random variable X. The function is:
$M_{X}(t)=\frac{1}{(1-2500t)^4}$

I am to determine the standard deviation.

I have no idea how to begin. I know that the standard form for a moment-generating function is:
$M_{X}(t)=E[e^{tx}]$

Neither my book nor Wikipedia give any clues as to what the variables mean. I certainly hope this is a situation where t (used in one explanation) = t (used in others). All I really know is, X is a random variable.

Should I just be trying to transform what I am given into the E[e^{tx} form?
Am I trying to determine what t is?
Is the form of the function I was given an indication of a specific type of continuous distribution? None of the distributions I've seen (uniform, normal, etc.) has a moment-generating function that looks like that.

2. Originally Posted by Boris B
In my problem, I have the moment-generating function on a random variable X. The function is:
$M_{X}(t)=\frac{1}{(1-2500t)^4}$

I am to determine the standard deviation.

I have no idea how to begin. I know that the standard form for a moment-generating function is:
$M_{X}(t)=E[e^{tx}]$

Neither my book nor Wikipedia give any clues as to what the variables mean. I certainly hope this is a situation where t (used in one explanation) = t (used in others). All I really know is, X is a random variable.

Should I just be trying to transform what I am given into the E[e^{tx} form?
Am I trying to determine what t is?
Is the form of the function I was given an indication of a specific type of continuous distribution? None of the distributions I've seen (uniform, normal, etc.) has a moment-generating function that looks like that.
The standard deviation is the (positive) square root of the variance. A formula for the variance is

$Var(X) = E(X^2) - [E(X)]^2$.

You can get these expected values from the moment generating function:

Theorem: $E(X^n) = \frac{d^n M_X(t)}{dt^n} \bigg{|}_{t=0}$.

Therefore:

To get E(X), differentiate the moment generating function once and then substitute t = 0.

To get E(X^2), differentiate the moment generating function twice and then substitute t = 0.

3. Originally Posted by mr fantastic
The standard deviation is the (positive) square root of the variance. A formula for the variance is

$Var(X) = E(X^2) - [E(X)]^2$.

You can get these expected values from the moment generating function:

Theorem: $E(X^n) = \frac{d^n M_X(t)}{dt^n} \bigg{|}_{t=0}$.

Therefore:

To get E(X), differentiate the moment generating function once and then substitute t = 0.

To get E(X^2), differentiate the moment generating function twice and then substitute t = 0.
Having re-read you original post, I think I should add the following:

E(X) means the expected (or mean) value of the random variable X.
E(X^2) means the expected (or mean) value of the random variable X^2.
etc.

4. Hello,

Nice ! I didn't know this way oO
Though it's strange to me

We learnt that $E(X)=\frac{dM_X(t)}{dt} \bigg{|}_{t={\color{blue} 1}}$

$E(X(X-1))=\frac{d^2 M_X(t)}{dt^2} \bigg{|}_{t=1}$

And so on...

5. Originally Posted by Moo
Hello,

Nice ! I didn't know this way oO
Though it's strange to me

We learnt that $E(X)=\frac{dM_X(t)}{dt} \bigg{|}_{t={\color{blue} 1}}$

$E(X(X-1))=\frac{d^2 M_X(t)}{dt^2} \bigg{|}_{t=1}$

And so on...
Hi Moo. The standard normal distribution provides a simple counter-example to what you've learnt:

The moment generating function is $m(t) = e^{t^2/2} \Rightarrow m'(t) = t e^{t^2/2} \Rightarrow m'({\color{red}0}) = 0 = \mu$.

But $m'({\color{red}1}) = \sqrt{e} \neq \mu$ ......

6. Is there a difference between the moment generating function of a random variable and the generating function of a random variable ?

7. Originally Posted by Moo
Is there a difference between the moment generating function of a random variable and the generating function of a random variable ?
The moment generating function of a random variable X is $m = E(e^{tx})$.

What's the definition of the generating function of a random variable X?

8. Actually, I don't know if it's the same in French...

What I call the generating function of a random variable is $E(t^X)=\sum_x t^x P(X=x)$

where x represents all the possible values X can take...

What does the moment generating function represent exactly ? In terms of probabilities I mean

9. Originally Posted by Moo
Actually, I don't know if it's the same in French...

What I call the generating function of a random variable is $E(t^X)=\sum_x t^x P(X=x)$

where x represents all the possible values X can take...

What does the moment generating function represent exactly ? In terms of probabilities I mean
Moment-generating function - Wikipedia, the free encyclopedia

Your generating function does not look too easy to calculate for continuous distributions .....? Do you know off-hand what it is for the standard normal distribution (I'm too lazy to calculate it)?

10. I will have difficulties to understand since there isn't the French page on wikipedia :/
But now I see where the difference is : mine is not for continuous distributions.
We'd use it for Poisson, Binomial, Geometric, etc... laws

11. I wonder if the two can be linked... For example if we know what the distribution for X is, it'd be possible to retrieve one from another.

What I still don't understand is this difference t=1 or t=0. Maybe it has to deal with a substitution when transforming P(X=x) into an integral...

Oh nevermind...I can't think about it right now

12. Originally Posted by Moo
I wonder if the two can be linked... For example if we know what the distribution for X is, it'd be possible to retrieve one from another.

What I still don't understand is this difference t=1 or t=0. Maybe it has to deal with a substitution when transforming P(X=x) into an integral...

Oh nevermind...I can't think about it right now
There's no big mystery.

The generating function g(t) is easily calculated for discrete probability distributions. For example:

For the geometric distribution:

$g(t) = \sum_{x = 1}^{\infty} t^x (1 - p)^{x - 1} p = p t \sum_{x = 1}^{\infty} [t (1 - p)]^{x - 1} = \frac{pt}{1 - t(1 - p)}$.

$g'(t) = \frac{p}{(t(p - 1) + 1)^2} \Rightarrow g'(1) = \frac{p}{p^2} = \frac{1}{p} = \mu$.

It's well known that the moment generating function is $m(t) = \frac{p e^t}{1 - (1 - p)e^t}$.

$m'(t) = \frac{p e^t}{e^t (p - 1) + 1)^2} \Rightarrow m'(0) = \frac{p}{p^2} = \frac{1}{p} = \mu$.

So $g'(1) = m'(0) = \mu$ - no big deal.

In fact, it's not hard to see that if $g(t) = \sum_x t^x f(x)$, then $g'(1) = \sum_x x f(x) = E(X)$.

13. I really have to read your link about the moment generating function, thanks again for it
Actually, I'm far too tired to think properly