laplace transform and conditioning Consider the continuous random variables $X, Z$ and $Y$. $Y = X + Z$. X and Z are independent. With probability $p$, $Z \sim X$, and with probability $1-p$, $Z = 0$.
My question is why I cannot get $Y = X + pX$?
What is the laplace transform of $Y$. I think it should be $\tilde{Y}(s) = p\tilde{X}(s)^2 + q\tilde{X}(s)$ but I am not sure.
 A: 
What is the laplace transform of $Y$. I think it should be $\tilde{Y}(s) = p\tilde{X}(s)^2 + q\tilde{X}(s)$ but I am not sure.

It is not; although the correct expression looks similar.

As Henry commented, for any given non-zero value for $X$, the value for $Y$ shall be either $X$ or $2X$ at probabilities of $(1-p)$ and $p$ respectively.   It will never be $(1+p)X$ … unless $p=1$ or $p=0$.
Now, the conditional expectation for $Y$ given $X$ shall be: $\mathsf E(Y\mid X)=(1+p)\,X$.  So indeed we obviously have:  $$\begin{align}\mathsf E(Y)&=\mathsf E(\mathsf E(Y\mid X))\\&=\mathsf E(pX+(1-p)X)\\&=(1+p)\,\mathsf E(X)\end{align}$$

However, you actually wish to evaluate $\tilde Y(s)$, the Laplace transform of $f_{\small Y}$ over its support; that is: $\mathsf E(\mathrm e^{-sY})$.  Still, you can find this in a similar manner:
$$\begin{align}\tilde Y(s)&=\mathsf E(\mathrm e^{-sY})\\&=\mathsf E(\mathsf E(\mathrm e^{-sY}\mid X))\\&~~\vdots\end{align}$$


 Note $\tilde X(s)=\mathsf E(\mathrm e^{-sX})$, so therefore $\mathsf E(\mathrm e^{-2sX})=\tilde X(2s)$.

