# 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.

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Covariance requires two random variables. You deal with one random variable $X = \sqrt{Y_1 \cdot Y_2}$. Did you mean to find variance of $X$ by chance ? – Sasha Apr 16 '12 at 22:07
sorry, i edited. In either case, I have to find expected value of X. – cap Apr 16 '12 at 22:12

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?

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+1 In fact, since $Y_1$ and $Y_2$ are identically distributed in this instance, it is necessary to compute only $E[Y_1]$ and $E[\sqrt{Y_1}]$. – Dilip Sarwate Apr 17 '12 at 1:16
@Dilip: Yes, I left that bit to the OP. I wanted to leave them some fun. :) – cardinal Apr 17 '12 at 1:16

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 $X$.

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Surely this is overkill since one can find $E[X]$ and $E[X^2]$ directly via the law of the unconscious statistician as $$E[X]=E[\sqrt{Y_1Y_2}]=\int_0^1\int_0^1\sqrt{y_1y_2}dy_1dy_2=\frac{4}{9}$$ and $$E[X^2]=E[Y_1Y_2]=E[Y_1]E[Y_2]=\frac{1}{4}$$ giving $\text{var}(X) = E[X^2]-(E[X])^2$?? – Dilip Sarwate Apr 17 '12 at 1:11