1. desperate...moment generating function

f(x)=3/64x^2(4-x)

what is the moment generating function?
how do i find the mean and variance?

any help would really be appreciated since i dont understand anything concerning this

thankx

2. Originally Posted by sasasa
f(x)=3/64x^2(4-x)

what is the moment generating function?
how do i find the mean and variance?

any help would really be appreciated since i dont understand anything concerning this

thankx
Do you mean $\displaystyle f(x)=\frac{3}{64}x^{2(4-x)}$ ?

3. Var(x) = $\displaystyle E(x^2) - E(x)^2$

$\displaystyle E(x) = M'(0)$

$\displaystyle E(x^2) = M''(0)$

MGF of f(X) = $\displaystyle E(e^{f(x)})$ so you'll have to integrate $\displaystyle e^{\frac{3}{64}x^{2(4-x)}}$ over a certain region.

4. Originally Posted by statmajor
MGF of f(X) = $\displaystyle E(e^{f(x)})$ so you'll have to integrate $\displaystyle e^{\frac{3}{64}x^{2(4-x)}}$ over a certain region.
This is not right. The moment generating function is defined as

$\displaystyle M_X (t) = \mathbb{E}e^{Xt} = \int_{\mathbb{R}} e^{xt}f(x) dx$

(for continuous distributions). I would help with the computation but I can't figure out what the density is supposed to be from the OP.

5. Originally Posted by sasasa
f(x)=3/64x^2(4-x)

what is the moment generating function?
how do i find the mean and variance?

any help would really be appreciated since i dont understand anything concerning this

thankx
I assume you mean $\displaystyle f(x) = \frac{3}{64} x^2 (4 - x)$. This pdf is incomplete as you do not include the interval over which this expression is defined (the support).

You should know that you need to calculate $\displaystyle E\left(e^{tX}\right) = \int e^{tx} \frac{3}{64} x^2 (4 - x) \, dx$ (and I have not included the integral terminals since you did not completely define the pdf). So please show all your working and clearly state where you are stuck.

6. Darn, can't believe I wrote that. Thanks for correcting me.

But the other parts of my post should be right.

Thanks again.

7. Sorry (so many functions)… so overwhelmed that i even posted the wrong function. The right one is:
f(x)=0.15e^(-0.15(x-0.5)), when x is bigger or equal to 0.5
Thanks again for all your precious help

8. Originally Posted by sasasa
Sorry (so many functions)… so overwhelmed that i even posted the wrong function. The right one is:
f(x)=0.15e^(-0.15(x-0.5)), when x is bigger or equal to 0.5
Thanks again for all your precious help
Can you set up the required integral using the definition of the moment generating function? Can you do the integration? Please show all that you can do and say where you're stuck.

(And please use normal size font so that we don't have to use microscopes to read it).

9. this is what i got:
Mx(t)=(-0.15e^(0.5t))/(t-0.15) but i am not sure if it is right

Can you please tell me if this is right:

Mean=E(X)=Mx´(0)=7.166

thanks

10. Originally Posted by sasasa
this is what i got:
Mx(t)=(-0.15e^(0.5t))/(t-0.15) but i am not sure if it is right

Can you please tell me if this is right:

Mean=E(X)=Mx´(0)=7.166

thanks
Your mgf is OK: - Wolfram|Alpha

I assume at this level you can correctly differentiate and substitute t = 0. You can check your own answer using the definition of E(X): Calculate $\displaystyle 0.15 \int_{1/2}^{+\infty} x e^{-0.15(x - 0.5)} \, dx$.

11. i think the mean is right...

did you get 0,25 as the variance?