If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), find (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].

Answers

Answer 1

Answer: Denote by $X_((n))$ the maximum order statistic, with $X_((n))=\max\{X_1,\ldots,X_n\}$ , and similarly denote by $X_((1))$ the minimum order statistic. Then the CDF for $X_((n))$ is

$F_{X_((n))}(x)=\mathbb P(X_((n))\le x)$

In order for there to be some $x$ that exceeds the value of $X_((n))$ , it must be true that $x$ exceeds the value of all the $X_i$ , so the above is equivalent to the joint probability

$F_{X_((n))}(x)=\mathbb P(X_1\le x,\ldots,X_n\le x)$

and since the $X_i$ are i.i.d., we have

$F_{X_((n))}(x)=\mathbb P(X_1\le x)\cdots\mathbb P(X_n\le x)=\mathbb P(X_1\le x)^n$
$\implies F_{X_((n))}(x)=F_X(x)^n$

where $X\sim\mathrm{Unif}(0,1)$ . We have

$F_X(x)=\begin{cases}0&\text{for }x<0\nx&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$

and so

$F_{X_((n))}(x)=\begin{cases}0&\text{for }x<0\nx^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((n))}(x)=\begin{cases}nx^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((n))]=\displaystyle\int_0^1xnx^(n-1)\,\mathrm dx=n\int_0^1x^n\,\mathrm dx=\frac n{n+1}$

Using similar reasoning, we can find the CDF for $X_((1))$ . We have

$F_{X_((1))}(x)=\mathbb P(X_((1))\le x)=1-\mathbb P(X_((1))>x)$
$F_{X_((1))}(x)=1-\mathbb P(X_1>x,\ldots,X_n>x)=1-\mathbb P(X_1>x)^n$
$F_{X_((1))}(x)=1-(1-\mathbb P(X\le x))^n=1-(1-F_X(x))^n$
$\implies F_{X_((1))}(x)=\begin{cases}0&\text{for }x<0\n1-(1-x)^n&\text{for }0\le x\le1\n1&\text{for }x>1\end{cases}$
$\implies f_{X_((1))}(x)=\begin{cases}n(1-x)^(n-1)&\text{for }0<x<1\n0&\text{otherwise}\end{cases}$
$\implies\mathbb E[X_((1))]=\displaystyle\int_0^1xn(1-x)^(n-1)\,\mathrm dx=\frac1{n+1}$

Answer 2

Answer:

Final answer:

The expected values of the maximum and minimum of independent and identically distributed (iid) uniform random variables, x1, x2, ..., xn, are given by E[max(x1, ..., xn)] = n / (n + 1) and E[min(x1, ..., xn)] = 1 / (n + 1) respectively.

Explanation:

In mathematics, particularly in probability theory and statistics, the question is related to independent and identically distributed (iid) random variables with a uniform distribution. The expected value or mean (E) of the maximum (max) and minimum (min) of these random variables is sought.

(a) The expected value of the max of 'n' iid uniform random variables, x1, x2, ..., xn, is calculated by integrating the nth power of x from 0 to 1. It can be found via the equation E[max(x1, ..., xn)] = n / (n + 1).

(b) Similarly, the expected value of the min of 'n' iid uniform random variables is acquired by doing (1 / (n + 1)). Hence, E[min(x1, ..., xn)] = 1 / (n + 1).

By understanding these, you could visualize the various outcomes of the random variables and their distributions, demonstrating how likely each outcome could occur.

Learn more about the Expected Value of Random Variables here:

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Answer:

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Answers

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