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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?

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Your general approach seems fine.

Note that your count below some cut-off point will have a binomial distribution so you're dealing with a binomial proportion (which you should already know how to compute a standard error for).

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?)

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  • $\begingroup$ I was certain that the variability would be zero but I lacked a formal justification for it. You just gave me that, it has a binomial distribution (didn't even think of that) I can work from here. Thank you!! $\endgroup$
    – Ramalho
    Oct 16 '16 at 12:13
  • $\begingroup$ The point I was making at the end there was you don't even need to know it's binomial as such; if the proportion is always going to be 1 then the proportion doesn't vary, by the definition of variance. $\endgroup$
    – Glen_b
    Oct 16 '16 at 20:52

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