Calculating MGF: Solutions for Undefined Limit Issue

[Usagi]

http://img253.imageshack.us/img253/7306/moments.jpg

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]

http://img253.imageshack.us/img253/7306/moments.jpg

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

$\chi$ $\sigma$

 

[Usagi]

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]

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

$\chi$ $\sigma$

 

[blue_raver22]

for bringing this issue to my attention. It appears that the MGF calculation provided in the image is incorrect. The correct approach would be to use the limit definition of the MGF, which is:

M(t) = \lim_{k\to\infty} E(e^{tX}) = \lim_{k\to\infty} \int_{-\infty}^{\infty} e^{tx} f(x) dx

Where f(x) is the probability density function of the random variable X. This approach should eliminate any undefined limits and provide a valid solution for the MGF.

In general, when dealing with undefined limits, it is important to carefully check the calculations and make sure that all steps are mathematically valid. It may also be helpful to consult with a colleague or refer to reliable sources to confirm the correct approach.