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Is there anyone could help to solve the following problem:

Suppose $\,h\left(x\right)\,$ is a known function and $\,y\left(x\right)\,$ is unknown, you may assume these two are nice functions. I am wondering how to get y here, say, express y by h without integral form? Actually, I have not got any clues to deal with it at this moment. So I use "nice" function, which means you can do any operation as you want to solve it.

$$y\left(s\right)=\int_0^{\infty} h\left(su\right)\, y\left(su\right)\, du$$

Considering a few people wants me to give more details: This question comes from an equation: $$Z=A(B+Z)$$ where you may consider A, B and Z are random variables and they are all independent. I am interested in what is the distribution of Z would look like. Then take the moment generating on both side ( You may want to use Laplace transform if you are worried about integrability. However，as I said, deal with nice function and you can omit it for a second.) and then take conditional probability on A on right hand side. This kind of integral is solvable when random variable has a finite range,say from a to b where a and b are finite numbers. But thing turns out to be difficult if $\infty$ is involved, so you will find: $$M_Z(s)=\int_0^\infty M_B(sa) M_Z(sa) dF_A(a)$$

and my question is how to get $M_Z(s)$.

Welcome any suggestions.

gouwangzhangdong
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1 Answers1

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If you substitute $v=su$ you get $$sy\left(s\right)=\int_{0}^{\pm\infty}h\left(v\right)y\left(v\right)dv$$ So $sy\left(s\right)$ is constant, but then $y\left(s\right)=c/s$ for a constant $c$. $$c=c\int_{0}^{\pm\infty}h\left(v\right)/vdv$$ which has a solution only if $$\int_{0}^{\pm\infty}h\left(v\right)/vdv=1$$

Joshua Tilley
  • 10,106