To use Rao-Blackwell, it is often a good idea to start with a naive estimator, then condition on the random part which is 'tricky'. In this case, if we know that $Y = y$, then the distribution of $X/Y\,|\, Y = y$ is normal, and we can compute its conditional expectation.
The naive Monte Carlo estimator of the above quantity can be computed by drawing $X_1,\ldots, X_n, Y_1, \ldots, Y_n$ independently from the standard normal, then computing.
$$
\hat{\delta} = \frac{1}{n}\sum_{i=1}^n I(X_i/Y_i\leq t).
$$
To get a better Rao-Blackwellized estimator, you can condition on the value of the $Y_i$'s. To start with we note that
$$
\mathbb{E}[I(X/Y \leq t)\,|\,Y = y] = \mathbb{P}(X\leq ty)I(y > 0) + \mathbb{P}(X\geq ty)I(y \leq 0) = \Phi(ty)I(y>0) + [1-\Phi(ty)]I(y\leq 0) = \Phi(t|y|),
$$
where $\Phi(\cdot)$ is the cdf. in the standard normal distribution. In the last inequality we have used the fact that $\Phi(-ty) = 1-\Phi(ty)$ by symmetry of the standard normal.
The Rao-Blackwellized estimator is then
$$
\hat{\delta}_{RB} = \mathbb{E}[\hat{\delta}\,|\, Y_1,\ldots, Y_n] = \frac{1}{n}\sum_{i=1}^n\mathbb{E}[I(X_i/Y_i\leq t)\,|\, Y_i] = \frac{1}{n}\sum_{i=1}^n \Phi(t|Y_i|).
$$
In particular, note from the final expression that there is actually no need to simulate $X_1, \ldots, X_n$ anymore, since we compute their conditional expectations.
We can then estimate the given probability as follows:
- Draw $Y_1,Y_2, \ldots, Y_n$ iid. standard normal
- Compute $\hat{\delta}_{RB} = \frac{1}{n}\sum_{i=1}^n \Phi(t|y_i|)$.
I advice you to try out the two approaches in repeated simulations! In particular, note that the Rao-Blackwellized estimator has much lower variance.
self-studytag. $\endgroup$