Answer to Question #180728 in Statistics and Probability for Nathan Drake

Question

(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) .

Answer 1

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

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

Then

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

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

We have:

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

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

Then

$\frac{{1 + a}}{{2a}} - \frac{{ - 1 + a}}{{2a}} = \frac{1}{2} \Rightarrow \frac{{1 + a + 1 - a}}{{2a}} = \frac{1}{2} \Rightarrow \frac{1}{a} = \frac{1}{2} \Rightarrow a = 2$

Let's find

$P(|X - 1| < 2) = P( - 2 < X - 1 < 2) = P( - 1 < X < 3) = F(3) - F( - 1) = 1 - \frac{{ - 1 + 2}}{4} = \frac{3}{4}$

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

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

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

$Var(X) = \frac{{{{\left( {b - a} \right)}^2}}}{{12}} = \frac{{{{\left( {3 - 0} \right)}^2}}}{{12}} = \frac{9}{{12}} = \frac{3}{4}$

If Y has exponential distribution with parameter α then

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

$Var(X) = Var(Y) \Rightarrow \frac{3}{4} = \frac{1}{{{\alpha ^2}}} \Rightarrow {\alpha ^2} = \frac{4}{3} \Rightarrow \alpha = \frac{{2\sqrt 3 }}{3}$

$P(Y > 0.5|Y < 1) = \int\limits_{0.5}^1 {\alpha {e^{ - \alpha x}}} dx = \int\limits_{0.5}^1 {\frac{{2\sqrt 3 }}{3}{e^{ - \frac{{2\sqrt 3 }}{3}x}}} dx = - \left. {{e^{ - \frac{{2\sqrt 3 }}{3}x}}} \right|_{0.5}^1 = - {e^{ - \frac{{2\sqrt 3 }}{3}}} + {e^{ - \frac{{\sqrt 3 }}{3}}} \approx 0.246$

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