Given the following joint pdf: $$f(x,y)=2x^{y-1}y^2e^{-y^2}, 0<x \leq 1, y \geq 0$$
We want to estimate $\theta=P\{X >\frac{Y}{Y+1}\}$. Consider $Z = I(X > \frac{Y}{Y+1})$. Generate a value of $g(Y) = E(Z|Y)$.
One approach seems to derive the marginal pdf $f(x)$ and then finding $\int_{\frac{y}{y+1}}^{\infty}f(x)dx$ since $E(Z|Y)=P(X > \frac{Y}{Y+1}|Y)$. Would this be correct? However, finding $f(x)$ seems tedious. I'm thinking using an importance sampling estimator may be a good approach. What do you think?
The desired probability can be computed by integrating the joint density over the region of interest: $$\begin{align*} \Pr[X > Y/(1+Y)] &= \int_{y=0}^\infty \int_{x=y/(1+y)}^1 2x^{y-1} y^2 e^{-y^2} \, dx \, dy \\ &= \int_{y=0}^\infty 2y^2 e^{-y^2} \left[\frac{x^y}{y}\right]_{x=y/(1+y)}^1 \, dy \\ &= \int_{y=0}^\infty 2y e^{-y^2} \left(1 - \frac{y^y}{(1+y)^y}\right) \, dy \\ \end{align*}$$ and at this point, it should be obvious that there is no closed form solution; however, numeric calculation is possible, giving $$0.46356908561532520568025428073320140197640792196629\ldots.$$ The integrand rapidly converges to $0$ as $y \to \infty$, so numeric evaluation proceeds easily.