Finding distribution by using mgf(moment generating function)

[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?

 

[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 f_Y is the pdf of Y. It seems to me that you first need to derive f_Y with brute force then substitute it in the MGF formula.

In general, f_Y 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.

 

[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...

 

[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?

 

[grimster]

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?

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.

 

[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
$$\alpha_r=\sum_{i=1}^{n}\alpha_i$$

 

[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.

 

[EnumaElish]

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.

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

 

[grimster]

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

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

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

 

[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.

 

[grimster]

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.

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