# Monte Carlo estimation of a proportion

I have been working on the following problem as a homework assignment.

Consider a random variable $$X$$ with pdf:

$$f(x) = 3x^2$$ if $$0\leq x \leq 1,\$$ else $$0$$

Let $$Y = h(X) = \sqrt{X^2+1}$$

Estimate $$P(Y < 1.5)$$ using the Monte Carlo method and also present the standard deviation of your estimate.

I started by calculating the cdf, $$C_X = x^3$$, then I applied the inverse transform and came to the conclusion that I could use $$X = U^{1/3}$$ to generate samples from the distribution of $$X$$.
I then proceeded to generate a $$10^5$$ samples of $$X$$ apply $$h(X)$$ such that I had samples from $$Y$$ and then simply counted how many where less than 1.5, finally I divided the counts by the number of samples ($$10^5$$).

Is my intuition correct? My estimate of the proportion is always 1 which I find very odd, where am I messing up?

Also.. on a purely analitical sense I did some manipulations:

$$P(Y \leq y) = P(h(X) \leq y) = P(\sqrt{X^2+1} \leq y) = P(X \leq \sqrt{y^2-1})=C_X(\sqrt{y^2-1})$$

Since $$0\leq x \leq 1$$ the only values $$y$$ can take are $$1 \leq y \leq \sqrt{2}$$ is that correct? If that's the case, since $$\sqrt{2} \leq 1.5$$ that can justify that $$P(Y < 1.5) = 1$$. Am I right?

I also have no idea on how I can calculate the SE of this estimate.

Can someone point me in the right direction?

If your question contains no errors, then you're correct that the $Y$ variable should never exceed $\sqrt{2}$, in which case the proportion would be 1. Before you try to compute the standard error, consider ... what will be the variability in your estimate of 1? (If you repeated it $k$ times what would be the distribution of the estimates of the proportion? / What would the variance be?)