Question #180728

(a) If X has a Uniform Distribution in (−𝑎, 𝑎), 𝑎 > 0 , find 𝑎 such that 𝑃(|𝑋| < 1) = 𝑃(|𝑋| > 1). Also find 𝑃(|𝑋 − 1| < 2) and 𝑃(|𝑋| > 2).

(b)If X has a uniform distribution in (0, 3) ant Y has exponential distribution with parameter α, find α that 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑌). Also find 𝑃(𝑌 > 0.5/𝑌 < 1) .


1
Expert's answer
2021-05-07T09:26:31-0400

(a) If X has a uniform distribution on a given segment, then the distribution function has the form:

F(x)={0,xax(a)a(a),a<x<a1,xa={0,xax+a2a,a<x<a1,xaF(x) = \left\{ {\begin{matrix} {0,\,\,x \le - a}\\ {\frac{{x - ( - a)}}{{a - ( - a)}},\,\, - a < x < a}\\ {1,\,\,x \ge a} \end{matrix}} \right.= \left\{ {\begin{matrix} {0,\,\,x \le - a}\\ {\frac{{x + a}}{{2a}},\,\, - a < x < a}\\ {1,\,\,x \ge a} \end{matrix}} \right.

Then

P(X<1)=P(1<X<1)=F(1)F(1)P(|X| < 1) = P( - 1 < X < 1) = F(1) - F( - 1)

P(X>1)=P(X>1)+P(X<1)=F()F(1)+F(1)F()=1F(1)+F(1)0=1F(1)+F(1)P(|X| > 1) = P(X > 1) + P(X < - 1) = F(\infty ) - F(1) + F( - 1) - F\left( { - \infty } \right) = 1 - F(1) + F( - 1) - 0 = 1 - F(1) + F( - 1)

We have:

F(1)F(1)=1F(1)+F(1)F(1) - F( - 1) = 1 - F(1) + F( - 1)

2F(1)2F(1)=1F(1)F(1)=122F(1) - 2F( - 1) = 1 \Rightarrow F(1) - F( - 1) = \frac{1}{2}

Then

1+a2a1+a2a=121+a+1a2a=121a=12a=2\frac{{1 + a}}{{2a}} - \frac{{ - 1 + a}}{{2a}} = \frac{1}{2} \Rightarrow \frac{{1 + a + 1 - a}}{{2a}} = \frac{1}{2} \Rightarrow \frac{1}{a} = \frac{1}{2} \Rightarrow a = 2

Let's find

P(X1<2)=P(2<X1<2)=P(1<X<3)=F(3)F(1)=11+24=34P(|X - 1| < 2) = P( - 2 < X - 1 < 2) = P( - 1 < X < 3) = F(3) - F( - 1) = 1 - \frac{{ - 1 + 2}}{4} = \frac{3}{4}

P(X>2)=P(X>2)+P(X<2)=F()F(2)+F(2)F()=11+00=0P(|X| > 2) = P(X > 2) + P(X < - 2) = F(\infty ) - F(2) + F( - 2) - F\left( { - \infty } \right) = 1 - 1 + 0 - 0 = 0

Answer: a=2, P(X1<2)=34P(|X - 1| < 2) = \frac{3}{4} , P(X>2)=0P(|X| > 2) = 0

(b) If X has a uniform distribution in (0, 3)  then

Var(X)=(ba)212=(30)212=912=34Var(X) = \frac{{{{\left( {b - a} \right)}^2}}}{{12}} = \frac{{{{\left( {3 - 0} \right)}^2}}}{{12}} = \frac{9}{{12}} = \frac{3}{4}

If Y has exponential distribution with parameter α then

Var(Y)=1α2Var(Y) = \frac{1}{{{\alpha ^2}}}

Var(X)=Var(Y)34=1α2α2=43α=233Var(X) = Var(Y) \Rightarrow \frac{3}{4} = \frac{1}{{{\alpha ^2}}} \Rightarrow {\alpha ^2} = \frac{4}{3} \Rightarrow \alpha = \frac{{2\sqrt 3 }}{3}

P(Y>0.5Y<1)=0.51αeαxdx=0.51233e233xdx=e233x0.51=e233+e330.246P(Y > 0.5|Y < 1) = \int\limits_{0.5}^1 {\alpha {e^{ - \alpha x}}} dx = \int\limits_{0.5}^1 {\frac{{2\sqrt 3 }}{3}{e^{ - \frac{{2\sqrt 3 }}{3}x}}} dx = - \left. {{e^{ - \frac{{2\sqrt 3 }}{3}x}}} \right|_{0.5}^1 = - {e^{ - \frac{{2\sqrt 3 }}{3}}} + {e^{ - \frac{{\sqrt 3 }}{3}}} \approx 0.246

Answer: α=233\alpha = \frac{{2\sqrt 3 }}{3} , P(Y>0.5Y<1)0.246P(Y > 0.5|Y < 1) \approx 0.246


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