Homework Help: Finding distribution by using mgf(moment generating function)

1. Sep 29, 2005

grimster

i have X_1,X_2,...X_n independant poisson-distributed variables with parameters: alfa_i and i=1,...k(unsure about this. however says so in the excercise)

i am supposed to find the distribution of
Y= SUM(from 1 to n) a_i*X_i where a_i>0

maybe one could use the "poisson paradigm" by thinking of each variable as a trial with p_i as the chance for success. so that

E[e^tX_i]=1+p_i(e^t - 1)

and

E[e^tX] is approximately
(product from i=1 to n) EXP{p_i(e^t - 1)

the problem is the a_i part. how do i find the mfg of Y?

2. Sep 29, 2005

EnumaElish

I'll go with MathWorld notation.

$$P_\nu(n)=\nu^n e^{-\nu}/n!$$

MGF of Y is defined as $m(t)=\sum_y e^{ty}f_Y(y)$ where fY is the pdf of Y. It seems to me that you first need to derive fY with brute force then substitute it in the MGF formula.

In general, fY will not be a Poisson pdf.

P.S. For large $\nu$ (that's your alfa BTW), $P_\nu$ can be approximated as a normal pdf with mean = standard dev. = $\nu$. If you use that approximation, then Y itself will be normal.

Last edited: Sep 30, 2005
3. Sep 30, 2005

grimster

got some help and this is what i have so far.

m(t; X_i) = exp[alpha_i*(exp(t)-1)].

The mgf of a_i*X_i is

m(t; a_i*X_i) = m(t*a_i; X_i) = exp[alpha_i*{exp(a_i*t)-1}].

The mfg of Y is

m(t; Y) = PROD[m(t; a_i*X_i)] = exp[SUM{alpha_i*(exp(a_i*t)-1)}].

the problem is now to say what distribution Y is....

4. Sep 30, 2005

EnumaElish

Excuse me, why isn't m(t; Y) = PROD[m(t; a_i*X_i)] = exp[SUM{alpha_i*(exp(a_i*t)-1)}] the answer to the problem?

5. Sep 30, 2005

grimster

i don't know. is the mgf also the distribution...?

the exercise asked us to find the distribution of Y, by finding the mgf of Y.

6. Sep 30, 2005

balakrishnan_v

$$E(e^{t\sum X_i})=\prod_{i}E(e^{t X_i})=e^{(e^t-1)(\lambda_1+..\lambda_n)}$$
Hence the resultant distribution is poisson with poisson parameter
$$\alpha_r$$
which is given by
$$\alpha_r=\sum_{i=1}^{n}\alpha_i$$

Last edited: Sep 30, 2005
7. Sep 30, 2005

EnumaElish

You left the weights $\alpha_i$ out of the definition of Y.

P.S. Balakrishnan, you have labeled the Poisson parameter once $\lambda$ and once $\alpha$. The $\alpha$ labels are prone to confusion as the OP used $\alpha$ for sum weights.

Last edited: Oct 1, 2005
8. Sep 30, 2005

EnumaElish

Excuse me, you're right. MGF is definitely not the pdf.

9. Oct 1, 2005

grimster

$$E(e^{tY})=\prod_{i}E(e^{ta_{i}X_{i}})=e^{\sum_{i}\left( e^{a_{i}t}-1\right) \left( \lambda _{i}\right) }$$

this is what i found the the MGF of Y to be. how do i know what the distribution of Y is?

10. Oct 1, 2005

EnumaElish

As I have posted above, I would use "brute force" (in Arabic, al jabr) to derive Y's pdf. Then see whether or how it also can be obtained from the MGF by comparing the MGF and the pdf formulas.

11. Oct 2, 2005

grimster

ok, but how do i do that then? how do i find the pdf of x_i*a_i ?