Variance of geometric mean of two uniform random variables Suppose $Y_1, Y_2$ are from a random sample of the uniform distribution on $[0, 1]$. How do I compute the variance of the geometric mean of two points in the interval $[0, 1]$?
I know the geometric mean is $X=(Y_1 Y_2)^{1/2}$.
I can't compute the variance since I'm not sure how to find the expected value of a radical.
 A: Hint: This is one of those instances where it helps to understand linearity of expectation and the product structure that occurs due to independence.
Recall that for an arbitrary random variable $X$ with finite second moment,
$$\newcommand{\E}{\mathbb E}
\mathrm{Var}(X) = \E X^2 - (\E X)^2 \>.
$$
In particular,
$$
\mathrm{Var}(\sqrt{Y_1 Y_2}) = \E Y_1 Y_2 - (\E\sqrt{Y_1 Y_2})^2 = (\E Y_1)(\E Y_2) - (\E \sqrt{Y_1})^2(\E \sqrt{Y_2})^2 \>,
$$
by independence of $Y_1$ and $Y_2$. (Why?)
Now, can you do the rest?
A: Let $X = \sqrt{Y_1 \cdot Y_2}$. Then, for $0\leqslant x\leqslant1$,
$$
   F_X(x) = \mathbb{P}(X \leqslant x) = \mathbb{P}(Y_1 Y_2 \leqslant x^2) = \mathbb{E}_{Y_2}( Y_1 Y_2 \leqslant x^2 | Y_2)) = \mathbb{E}\left( \min\left(1, \frac{x^2}{Y_2} \right) \right) = \int_0^{x^2} 1 \mathrm{d} y + \int_{x^2}^1 \frac{x^2}{y} \mathrm{d} y
$$
You can now compute the probability distribution function $f_X(x)$ by differentiating, and then compute variance by evaluating first and second moment of $X$.
