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One of the problems of my Statistics assignment requires me to calculate the expected value of a minimum. Here's the situation:

The following density function is given: $f(x) = \frac{\theta}{x^2}$ where $x\ge\theta$ and $\theta>0$

I have to calculate: E[min{$X_i$}]

My initial guess was that the smallest possible value of $X$ is $\theta$ sense $x\ge\theta$ so the expected value of the minimum would be $\theta$, but then again, in an acquired sample, you can't be 100% certain that the smallest possible value of $X$ will be one of the observations. So what then?

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Minima and expectations of nonnegative random variables are both well suited to complementary CDF, two remarks which together make for a painless solution.

In the present case, $\mathrm P(X_i\geqslant x)=\theta/x$ for every $i$ and every $x\geqslant\theta$ hence $M_n=\min\limits_{1\leqslant i\leqslant n}X_i$ is such that $\mathrm P(M_n\geqslant x)=\mathrm P(X_1\geqslant x)^n$ is $(\theta/x)^n$ if $x\geqslant\theta$ and $1$ if $x\leqslant\theta$.

Now, $\mathrm E(Y)=\int\limits_0^{+\infty}\mathrm P(Y\geqslant x)\mathrm dx$ for every nonnegative random variable $Y$, hence $$ \mathrm E(M_n)=\int_0^\theta 1\cdot\mathrm dx+\int_\theta^{+\infty}(\theta/x)^n\mathrm dx=\theta+\theta\int_1^{+\infty}\mathrm dx/x^n=\theta+\theta/(n-1), $$ that is, $$ \mathrm E(M_n)=n\theta/(n-1). $$ One sees that $M_1=X_1$ is not integrable, that $M_n$ is integrable for every $n\geqslant2$, and that $M_n\to\theta$ almost surely and in $L^1$ when $n\to\infty$.

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Let's compute the PDF of the minimum: $$\phi(t) = P(\min \leq t) = 1 - P(\min \gt t) = 1 - P(\forall_i X_i \gt t) = 1 - \prod_i P(X_i > t) = \\ = 1 - P(X_1 \gt t)^n.$$

So the density $g$ of the minimum is $\frac{d\phi}{dt}$: $$g(t) = \phi'(t) = - n \cdot P(X_1 \gt t)^{n-1} \cdot P(X_1 \gt t)' = - n \cdot P(X_1 \gt t)^{n-1} \cdot (-f(t)) = \\ = n \cdot P(X_1 \gt t)^{n-1} \cdot f(t)$$

The rest is just computing $P(X_1>t)$ which is simple integration and then finding $\mathbb{E}\min=\frac{\int t \cdot g(t)}{\int g(t)}$ by another integration.

I hope that helps:)

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In the OP's case, however the function is not integrable and the expected value is infinite. – Byron Schmuland Feb 25 '12 at 17:54
@ByronSchmuland: Are you sure? I think it is not integrable for $n = 1$ but integrable for $n=2,3,4,5,\ldots$. – savick01 Feb 25 '12 at 18:00
Oh, you are quite right. – Byron Schmuland Feb 25 '12 at 18:10
I think it is particularly nice to replace $P(X_1 > t)$ with $1-F(t)$, the CDF. – FooBar Apr 24 '15 at 21:20

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