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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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Answer:

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Step-by-step explanation:

Using the standard equation of a circle with center at (h, k) and radius r:

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Alternate Hypothesis, H_A : \mu\neq 200 clicks a day

Step-by-step explanation:

We are given that a website developer wished to analyze the clicks per day on their newly updated website.

The website developer wants to know if the number of clicks per day is different than 200 clicks a day, on average.

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