Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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 $n\to\infty$.

share|improve this answer

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 $\mathbb{E}\min=\int t \cdot g(t)$ by another integration.

I hope that helps:)

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.