Question about convolution of two discrete random variable

There is poisson random variable X with T*lambda_MUE.

There is poisson random variable Z with lambda_SMC.

And i define new random variable K=aX+bZ. (a and b is positive integer which is greater than 0)

To calculate pmf of K i used conv function.

And calculate using below command but it is not equal to E[K]

What i expect is 123 but result is 43 .

What is wrong?

Thank you

lambda_MUE = 2;
lambda_SMC = 3;
alpha = 3; 
beta = 1; 
T=20; 
g=alpha*(0:1:110);
z=beta*(0:1:110);
K=0:1:(max(g)+max(z));
P_X = ((T*lambda_MUE).^(g/alpha))./(factorial(g./alpha)).*exp(-T*lambda_MUE);
P_Z = (lambda_SMC.^(z./beta))./(factorial(z./beta)).*exp(-lambda_SMC);
P_K = conv(P_X,P_Z);
sum(K(1:1:length(P_K)).*P_K)

[Torsten]

20*2+1*3=43

How do you arrive at 123 ?

Best wishes

Torsten.

[h kim]

I expect below

T*lambda_MUE*alpha + lambda_SMC*beta=20*2*3+1*3=123

[Torsten]

No. You form the convolution of two Poisson random variables with expected values T*lambda_MUE and lambda_SMC. The expected value of their sum is T*lambda_MUE + lambda_SMC = 43.

Best wishes

Torsten.

[h kim]

There is mistake.

I written X and Y are poisson on outside code.

But in code P_X and P_Y is not poisson sorry about that.

Please check below page

P_X and P_Y is not poisson.

http://math.stackexchange.com/questions/1640151/what-is-distribution-of-poisson-multiplied-by-positive-constant

[Torsten]

X and Y in your code are Poisson random variables since g/alpha = (0:1:110) and z/beta = (0:1:110).

Best wishes

Torsten.

[h kim]

But support of X is (0,alpha,2alpha,......) so expectation value of X is not lambda*T but alpha*lambda*T. And that of Y is (0, beta,2beta,....). Expectation value of P_X is 120 and that of P_Y is 3. Therefore i expect expectation value of Z is 123.

[Torsten]

I don't see any line in your code where the above comes into play.

All the calculations assume that the support for X and Y is 0,1,2,... .

Best wishes

Torsten.

[h kim]

Oh.... i incorrectly think.

Thank you i solve!

I really appreciate you!

[the cyclist]

The best form of thanks is to upvote and/or accept an answer that was helpful. This rewards the contributor, and may help guide future users trying to solve similar problems.

[John D'Errico]

Of course, where there is no explicit answer, it reflects a problem with the site, that we cannot up-vote comments.

[the cyclist]

Indeed. (I'm so used to making that comment, that I failed to notice there was no answer here!)

[John BG]

we are all glad you found out, can we now focus on the question?

Answer 1

[h kim]

If we use above code, P_X describe probability density in support [0,1,2,3.......]

But actual aim is probability density function describing probability density in support [0,alpha,2alpha,.....]

So we need additional code which make zero probability density when support is not multiple of alpha like below.

-----------------------------------------------------------------------------

for i=0:1:300

    if(mod(i,alpha)==0)

        P_X=[P_X ((lambda_MUE).^(i/alpha))./(factorial(i./alpha)).*exp(-lambda_MUE)];

    else

        P_X=[P_X 0];

    end

end

-----------------------------------------------------------------------------