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Let $\ X_1,X_2,X_3$ be an independent continuous random variables that are uniformly distributed on $\ [0, n]$.

Let $\ M = M=max⁡{(X_1,X_2,X_3)}$

$\ P(M≤X)=?$ for general $\ X∈[0,n]$

$\ E[M]=?$

I Know that $\ P(X≤t)=F(t)=(t-a)/b=t/n$
$\ P(M≤X)=P(X>M)=1-F(t)=1-M/n$ , Is that true?
How I calculate $\ E[M]=?$

Assia Khteb
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2 Answers2

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To find $P(M\le t)$ :

Calculate it by definition : $P(M\le t) = P(Max\{X_1,X_2,X_3 \} \le t) = P(\cap_{i=1}^3 (X_i\le t) )$ now use independence and finish.

For $E[M]$ , again calculate it by definition.

Differentiate $F_M(t) = P(M\le t)$ to get $f_M(t)$ and then find $E[M] = \int_\Bbb R t f_M(t) dt$.

infinity
  • 946
  • $\ a≤X≤b $ $\ P(X_1≤X)=P(a≤X_1≤b)=F(b)-F(a)= (b-a)/(b-a)-(a-a)/(b-a)=1 $ Therefore , $\ P(M≤X)=111=1 $ $\ E[M]=∫(X1)=X^2/2$ I’m not sure about the answer* @infinity – Assia Khteb Jan 28 '20 at 22:12
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Hints:

  • $P(M \le x) = P(X_1 \le x, X_2 \le x, X_3 \le x) = P(X_1 \le x) P(X_2 \le x) P(X_3 \le x)$
  • Because $M$ is nonnegative you can use the tail sum formula for expectation: $E[M] = \int_0^n P(M \ge t) \, dt$. You can use the result of the previous question to find the integrand.
angryavian
  • 89,882
  • $\ a≤X≤b $ $\ P(X_1≤X)=P(a≤X_1≤b)=F(b)-F(a)=(b-a)/(b-a)-(a-a)/(b-a)=1 $ Therefore , $\ P(M≤X)=111=1 $ – Assia Khteb Jan 28 '20 at 22:04