I'm not totally sure how to even word this question, but I need to find the first and second moments of two variables, $M$ and $N$ as defined by:

$M=\min(X_1,X_2,\dots,X_n)$ and $N=\max(X_1,X_2,\dots,X_n)$ where $(X_1,X_2,\dots,X_n)$ is uniformly distributed on the interval $(0,1)$.

Here's what we have:





I'm a little stuck actually computing the expectations, though. By definition, I know


But I'm getting really confused by the actual computation for my specific purposes of calculating $E[M], E[M^2], E[N], E[N^2]$. Thanks for any guidance.


If you were given\taught the Beta function, then, since you know the density of $M$ (the minimum), \begin{align*} E[M] &= \int_0^1 mf_M(m)\,dm\\ &=\int_0^1m\cdot n(1-m)^{n-1}\,dm\\ &=n \int_0^1\cdot m^{2-1}(1-m)^{n-1}\,dm\\ &=n B(2,n)\\ &=n \cdot\frac{(2-1)!(n-1)!}{(2+n-1)!}\\ &=\frac{n!}{(n+1)!}\\ &=\frac{1}{n+1}. \end{align*} Similarly, \begin{align*} E[M^2] &= \int_0^1m^2f_M(m)\,dm\\ &=\int_0^1m^2\cdot n(1-m)^{n-1}\,dm\\ &=n \int_0^1m^{3-1}(1-m)^{n-1}\,dm\\ &=nB(3,n)\\ &=n\cdot\frac{(3-1)!(n-1)!}{(3+n-1)!}\\ &=\frac{2!n!}{(n+2)!}\\ &=\frac{2}{(n+1)(n+2)} \end{align*} You can verify that $$\text{Var}[M] = E[X^2]-\{E[X]\}^2 = \frac{n}{(n+1)^2(n+2)}.$$

This will work $N$ also.

  • $\begingroup$ So the moments of $M$ and $N$ are the same? $\endgroup$ – Taylor Jan 27 '16 at 5:52
  • $\begingroup$ @Taylor No, $E[N] = \frac{n}{n+1}$. $\endgroup$ – Em. Jan 27 '16 at 5:54
  • $\begingroup$ Is the other answer on this page wrong, then? $\endgroup$ – Taylor Jan 27 '16 at 6:01
  • $\begingroup$ @Taylor Not completely. We actually calculated the same thing. I did not proofread that person's work until a moment ago. But it is right, if you switch the $N$ to an $M$. $\endgroup$ – Em. Jan 27 '16 at 6:03
  • $\begingroup$ @Taylor You will learn that distribution of $M$ follows a $\text{Beta}(1,n)$, so that is how I checked my work. Since its properties are know, then I can check its expectation and variance. $\endgroup$ – Em. Jan 27 '16 at 6:05

Here I gave you explicitely forms of each distribution.

For $N$:

$\begin{eqnarray}E[N]&=&\int_0^1nx(1-x)^{n-1}dx\\ &=&\frac{n}{n^2+n}\\ &=&\frac{1}{n+1} \end{eqnarray}$

$\begin{eqnarray}E[N^2]&=&\int_0^1nx^2(1-x)^{n-1}dx\\ &=&\frac{2n}{n(n+1)(n+2)}\\ &=&\frac{2}{(n+1)(n+2)} \end{eqnarray}$

Can you do for $M$?

Hint $$E[M]=\int_0^1 {n\cdot x\cdot x^{n-1}} dx\mbox{ and }E[M^2]=\int_0^1 {n\cdot x^2\cdot x^{n-1}} dx$$

  • $\begingroup$ Do you mind explaining the subscripts in the explicit forms? $\endgroup$ – Taylor Jan 27 '16 at 4:39
  • $\begingroup$ It means indicator functions: $1_{A}(t)=1$ if $t\in A$ and $0$ in other case, and have nice properties: if $A$ and $B$ are disjoints, then $(f(x)1_A(x)+g(x)1_B(x))^n=f(x)^n1_A(x)+g(x)^n1_B(x)$. In particular $[x1_{A}(x)]^n=x^n1_{A}(x)$ (that is, the exponents don't affect the indicator function) $\endgroup$ – sinbadh Jan 27 '16 at 4:42
  • $\begingroup$ Is that necessary? $\endgroup$ – Taylor Jan 27 '16 at 4:47
  • $\begingroup$ If $Z\in U(0,1)$, then $f_Z(z)=\left\{\begin{array}{cc}\frac{1}{b-a}&\mbox{if }z\in[a,b]\\0&\mbox{i.o.c.}\end{array}\right.$, ok. But using indicator functions it writes only like $f_Z(z)=\frac{1}{b-a}1_{[a,b]}(z)$. Then, a first use is compact notations. $\endgroup$ – sinbadh Jan 27 '16 at 5:02
  • $\begingroup$ I calculated $E[M]=\frac{n}{2}$ and $E[M^{2}]=\frac{n}{3}$. Is this correct? $\endgroup$ – Taylor Jan 27 '16 at 5:39

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