0
$\begingroup$

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:

$F_N(m)=1-(1-F_X(m))^n$

$f_N(m)=n(1-F_X(m))^{n-1}f_X(m))$

$F_M(m)=F_x(m)^n$

$f_M(m)=nF_X(m)^{n-1}f_X(m)$

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

$E[X^k]=\int_{0}^{1}x^kf_X(x)dx$,

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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