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goblinko [34]
3 years ago
12

If x1, x2, . . . , xn are independent and identically distributed random variables having uniform distributions over (0, 1), fin

d (a) e[max(x1, . . . , xn)]; (b) e[min(x1, . . . , xn)].
Mathematics
1 answer:
sveta [45]3 years ago
5 0
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 }x1\end{cases}

and so

F_{X_{(n)}}(x)=\begin{cases}0&\text{for }x1\end{cases}
\implies f_{X_{(n)}}(x)=\begin{cases}nx^{n-1}&\text{for }0
\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 }x1\end{cases}
\implies f_{X_{(1)}}(x)=\begin{cases}n(1-x)^{n-1}&\text{for }0
\implies\mathbb E[X_{(1)}]=\displaystyle\int_0^1xn(1-x)^{n-1}\,\mathrm dx=\frac1{n+1}
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In this question:

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