Stochastic ordering when multiplying pdfs

Let $X$ and $Y$ be independent random variables with probability density functions $f_X$ and $f_Y$ respectively. Let $X$ first-order stochastically dominate $Y$.

Consider a random variable $W$ whose probability density function is the product of the two pdfs, that is, $f_W(x) = \frac{1}{N}f_X(x) f_Y(x)$, where N is a normalization factor $\int_{-\infty}^{\infty}f_X(x) f_Y(x)dx$.

I want to prove that $W$ falls "in between" $X$ and $Y$, that is:

1. $X$ first-order stochastically dominate $W$.
2. $W$ first-order stochastically dominates $Y$.

Is this statement true?

Take the example

• $X \sim \text{Exp}(1)$ i.e. $f_X(x)=e^{-x}$ for $x\ge 0$
• $Y \sim \text{Exp}(2)$ i.e. $f_Y(x)=2e^{-2x}$ for $x\ge 0$

Then $f_X(x) f_Y(x) = 2e^{-3x}$ for $x\ge 0$ giving $N=\frac23$

so $W \sim \text{Exp}(3)$ i.e. $f_W(x)=3e^{-3x}$ for $x\ge 0$

$W$ does not stochastically dominate $Y$ here; instead $Y$ stochastically dominates $W$