Calculating the MGF

Usagi

Member

This a pretty weird question... because:

$$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$$

But the limit: $$\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$$ is undefined?

How am I meant to compute the MGF then?

Thanks

chisigma

Well-known member

This a pretty weird question... because:

$$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$$

But the limit: $$\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$$ is undefined?

How am I meant to compute the MGF then?

Thanks
If $\displaystyle 1-t>0 \implies t<1$ is...

$\displaystyle E\{e^ {t\ X}\}= \int_{0}^{\infty} e^{-x\ (1-t)}\ dx = - |\frac{e^{-x\ (1-t)}}{1-t}|_{0}^{\infty} = \frac{1}{1-t}$ (1)

The condition $t<1$ is no limitation because pratically we are interested to the function $M(t)$ and its derivatives in $t=0$...

Kind regards

$\chi$ $\sigma$

Last edited:

Usagi

Member
Thanks chisigma,

However how did you know to set t-1>0? I thought the restriction on t was that there exists a positive b, such that $$t \in (-b,b)$$

How does that relate with setting t-1>0 though?

Thanks again

chisigma

Well-known member
Thanks chisigma,

However how did you know to set t-1>0? I thought the restriction on t was that there exists a positive b, such that $$t \in (-b,b)$$

How does that relate with setting t-1>0 though?

Thanks again
The integral defining the moment generating function...

$\displaystyle M(t)= E\{e^{t\ X}\}= \int_{0}^{\infty} e^{-x\ (1-t)}\ dx$ (1)

... converges for $\displaystyle t<1$ to $\displaystyle M(t)= \frac{1}{1-t}$. The series expansion...

$\displaystyle M(t)= \frac{1}{1-t}= \sum_{n=0}^{\infty} t^{n}$ (2)

... converges for $\displaystyle -1<t<1$ and (2) allows You an easily computation of the moments...

$\displaystyle E\{X^{n}\}= M^{(n)}(0)= n!$ (3)

Kind regards

$\chi$ $\sigma$