Probability Distribution of R.V r: E_m[m exp(-rm)]

[gkannan16]

Homework Statement

Assuming 'm' is deterministic the probability distribution of a Random Variable(R.V) r is f(r)=m exp(-rm) Now m itself is a another R.V with a probability distribution g(m). Is it correct to say that now the probability distribution of 'r' is f(r)=E_m [m exp(-rm)] where E_m is the statistical expectation operation with respect to 'm'. If it is correct can some one give me a mathematical reference (some journal publications or book)?

Homework Equations

f(r)=m exp(-rm)

The Attempt at a Solution

 

[Redbelly98]

Homework Statement

Assuming 'm' is deterministic the probability distribution of a Random Variable(R.V) r is f(r)=m exp(-rm) Now m itself is a another R.V with a probability distribution g(m). Is it correct to say that now the probability distribution of 'r' is f(r)=E_m [m exp(-rm)] where E_m is the statistical expectation operation with respect to 'm'. If it is correct can some one give me a mathematical reference (some journal publications or book)?

Homework Equations

f(r)=m exp(-rm)

The Attempt at a Solution

That doesn't look right. The final f(r), after accounting for the distribution of m's, should not depend on m.

I think you need to weight the first f(r) (the one that does depend on m) by g(m), then integrate that over m to get the final f(r).

 

[gkannan16]

Thanks a lot for your reply. I agree with you that the final f(r), after accounting for the distribution of m, should not depend on m. Now shall i follow these steps

1. First find f(r) as a function of r and m where m is a random variable with the distribution g(m).
After that
2. Now int_{range of m}f(r)g(m)dm to get rid of m and find the final expression for f(r)

Are these steps correct?