Expected Value 5

This approach may be long, but I usually like to use it in solving continuous probability problems.

Divide our interval $[-1,1]$ into $m$ sub-intervals ( $m$ is very large) each of width $\frac{2}{m}$ and to be considered as a single value $-1+2\frac{a}{m}$ , where $a$ can take the values $\{1,2,...,m\}$ .

Now , let $\{A_{i}\}_{1}^n$ be $n$ numbers randomly(and independently) chosen from the given interval, and $X$ be the random variable corresponding to the minimum chosen number.

Therefore:

$\displaystyle A_{i} = -1+2\frac{a_{i}}{m} \;\;\;\;\;\;\;\;\; \text{ where each variable } \;\;\; a_1,a_2,...,a_n\;\;\; \text{ is randomly chosen from } \;\;\; \{1,2,...,m\}$

and

$\displaystyle X \;\;\;\text{can take values of}\;\;\; -1+2\frac{k}{m} \;\;;\;\; k\in\{1,2,...,m\}$

Note that for every integer $1\leq j \leq m$ :

$\displaystyle A_j \;\; \text{is minimum among }\;\{A_i\}_1^n\;\;\;\;\;<=> \;\;\;\;\; a_j \;\;\text{is minimum among }\;\;\{a_i\}_1^n$

Therefore we have:

$\displaystyle P\big(X=-1+2\frac{k}{m}\big)$

$\displaystyle =P(a_1=k\;\;\;\&\;\;\; a_i\geq k\;\;\forall i) +P(a_2=k \;\;\;\&\;\;\; a_i\geq k\;\;\forall i)+...+P(a_n=k \;\;\;\&\;\;\; a_i\geq k\;\;\forall i)-\color{#D61F06}{(n-1)P(a_i=k\;\;\forall i)}$

(Obviously,the non-red terms are equal)

$\displaystyle = nP(a_1=k\;\;\;\&\;\;\; a_i\geq k\;\;\forall i) -\color{#D61F06}{(n-1)P(a_i=k\;\;\forall i)}$

And since the $a_i$ 's are independent,and that $\boxed{a_1=k\;\;\;\&\;\;\; a_1\geq k\;\;<=>\;\;a_1=k\;}$ :

$\displaystyle =nP(a_1=k)P(a_2\geq k)P(a_3\geq k)...P(a_n\geq k)-\color{#D61F06}{(n-1)P(a_1=k)P(a_2=k)P(a_3=k)...P(a_n=k)}$

$\displaystyle =n.\frac{1}{m}.\frac{m-k+1}{m}.\frac{m-k+1}{m}...\frac{m-k+1}{m}-\color{#D61F06}{(n-1)\frac{1}{m}.\frac{1}{m}.\frac{1}{m}...\frac{1}{m}}$

$\displaystyle =\frac{n}{m} \bigg(\frac{m-k+1}{m}\bigg)^{n-1}-\frac{n-1}{m^n}$

Hence:

$\displaystyle E(X^2)=\sum_{k=1}^{m} \big(-1+2\frac{k}{m}\big)^2\Bigg(\frac{n}{m} \bigg(\frac{m-k+1}{m}\bigg)^{n-1}-\frac{n-1}{m^n}\Bigg)$

$\displaystyle = n.\frac{1}{m}\sum_{k=1}^{m} \big(-1+2\frac{k}{m}\big)^2\Bigg(\bigg(\frac{m+1}{m}-\frac{k}{m}\bigg)^{n-1}-\frac{n-1}{m^n}\Bigg)$

Therefore:

$\displaystyle E_n=\lim_{m\to \infty} E(X^2)=n\int_0^1 (-1+2x)^2\big((1-x)^{n-1}-0\big)dx$

The integral is easy to solve, just use the substitution $u=1-x$ ,then expand and integrate to obtain at last:

$\displaystyle \boxed{E_n=\frac{n^2-n+2}{n^2+3n+2}}$

Now, substituting the result in the desired expression to be calculated, we obtain:

$\displaystyle P=\prod_{n=1}^{\infty} \frac{n^2+3n-2}{n^2+3n+2}$

$\displaystyle =\prod_{n=1}^{\infty} \frac{(n-a)(n+3+a)}{(n+1)(n+2)}$

( $a$ is a root of the upper polynomial,so is $\;-3-a\;$ since their sum is $-3$ .)

$\displaystyle =\frac{\prod_{n=1}^{\infty} (n-a)\color{#20A900}{\prod_{n=1}^{\infty} (n+3+a)}}{\color{#D61F06}{\prod_{n=1}^{\infty} (n+1)}\color{#3D99F6}{\prod_{n=1}^{\infty} (n+2)}}$

$\displaystyle =\frac{\prod_{n=1}^{\infty} (n-a)\color{#20A900}{\frac{1}{1+a}\frac{1}{2+a}\frac{1}{3+a}\prod_{n=1}^{\infty} (n+a)}}{\color{#D61F06}{\frac{1}{1}\prod_{n=1}^{\infty} (n)}\color{#3D99F6}{\frac{1}{1}\frac{1}{2}\prod_{n=1}^{\infty} (n)}}$

$\displaystyle =\frac{2}{(a+1)(a+2)(a+3)}\prod_{n=1}^{\infty}\big(1-\frac{a^2}{n^2}\big)$

Now , the above infinite product alone is equal to $\;\frac{\sin (\pi a)}{\pi a}\;$ ,which is a well known formula and can also be derived easily using the Gamma Function and its Reflection Formula. The value of $a$ here is any root of the polynomial $\;n^2+3n-2\;$ . So after applying simple trigonometry and some manipulations, we get the desired answer:

$\displaystyle \boxed{\frac{3}{4\pi}\cos\big(\frac{\pi \sqrt{17}}{2}\big)}$

Hints:

$E_n = \frac{n^2-n+2}{n^2+3 n+2}=1-\frac{4 n}{n^2+3 n+2}$ (prove this)

Use sin/cos infinite product formulas to compute:

$\displaystyle\prod _ {n = 2}^{\infty}\left (1 - \frac {4} {n^2 + 3 n + 2} \right)$

Edit

Here's how to find $E_n$ . It's pretty simple. First note that $E_n$ is also equal to the expected value of $Y^2$ , where Y is the random variable of the largest of the n numbers (since $[-1,1]$ is symmetric about the origin).

Suppose the interval is $[0,1]$ not $[-1,1]$ . Note that the probability of n-1 of the numbers to be less than x is $x^{n-1}$ . Hence translating and dilating to the interval $[-1,1]$ , the cdf of Y is $(x/2+1/2)^{n-1}$ . So

$E_n = \frac{\displaystyle\int_{-1}^1 x^2 \left(\frac{x}{2}+\frac{1}{2}\right)^{n-1} \, dx}{\displaystyle\int_{-1}^1 \left(\frac{x}{2}+\frac{1}{2}\right)^{n-1} \, dx}$

(We have to normalize.)

Anybody else have another way to do it?