(a) If X has a uniform distribution on a given segment, then the distribution function has the form:
F ( x ) = { 0 , x ≤ − a x − ( − a ) a − ( − a ) , − a < x < a 1 , x ≥ a = { 0 , x ≤ − a x + a 2 a , − a < x < a 1 , x ≥ a 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. F ( x ) = ⎩ ⎨ ⎧ 0 , x ≤ − a a − ( − a ) x − ( − a ) , − a < x < a 1 , x ≥ a = ⎩ ⎨ ⎧ 0 , x ≤ − a 2 a x + a , − a < x < a 1 , x ≥ a
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 ( − 1 < X < 1 ) = F ( 1 ) − F ( − 1 )
P ( ∣ X ∣ > 1 ) = P ( X > 1 ) + P ( X < − 1 ) = F ( ∞ ) − F ( 1 ) + F ( − 1 ) − F ( − ∞ ) = 1 − F ( 1 ) + F ( − 1 ) − 0 = 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) P ( ∣ X ∣ > 1 ) = P ( X > 1 ) + P ( X < − 1 ) = F ( ∞ ) − F ( 1 ) + F ( − 1 ) − F ( − ∞ ) = 1 − F ( 1 ) + F ( − 1 ) − 0 = 1 − F ( 1 ) + F ( − 1 )
We have:
F ( 1 ) − F ( − 1 ) = 1 − F ( 1 ) + F ( − 1 ) F(1) - F( - 1) = 1 - F(1) + F( - 1) F ( 1 ) − F ( − 1 ) = 1 − F ( 1 ) + F ( − 1 )
2 F ( 1 ) − 2 F ( − 1 ) = 1 ⇒ F ( 1 ) − F ( − 1 ) = 1 2 2F(1) - 2F( - 1) = 1 \Rightarrow F(1) - F( - 1) = \frac{1}{2} 2 F ( 1 ) − 2 F ( − 1 ) = 1 ⇒ F ( 1 ) − F ( − 1 ) = 2 1
Then
1 + a 2 a − − 1 + a 2 a = 1 2 ⇒ 1 + a + 1 − a 2 a = 1 2 ⇒ 1 a = 1 2 ⇒ a = 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 2 a 1 + a − 2 a − 1 + a = 2 1 ⇒ 2 a 1 + a + 1 − a = 2 1 ⇒ a 1 = 2 1 ⇒ a = 2
Let's find
P ( ∣ X − 1 ∣ < 2 ) = P ( − 2 < X − 1 < 2 ) = P ( − 1 < X < 3 ) = F ( 3 ) − F ( − 1 ) = 1 − − 1 + 2 4 = 3 4 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 − 1∣ < 2 ) = P ( − 2 < X − 1 < 2 ) = P ( − 1 < X < 3 ) = F ( 3 ) − F ( − 1 ) = 1 − 4 − 1 + 2 = 4 3
P ( ∣ X ∣ > 2 ) = P ( X > 2 ) + P ( X < − 2 ) = F ( ∞ ) − F ( 2 ) + F ( − 2 ) − F ( − ∞ ) = 1 − 1 + 0 − 0 = 0 P(|X| > 2) = P(X > 2) + P(X < - 2) = F(\infty ) - F(2) + F( - 2) - F\left( { - \infty } \right) = 1 - 1 + 0 - 0 = 0 P ( ∣ X ∣ > 2 ) = P ( X > 2 ) + P ( X < − 2 ) = F ( ∞ ) − F ( 2 ) + F ( − 2 ) − F ( − ∞ ) = 1 − 1 + 0 − 0 = 0
Answer: a=2, P ( ∣ X − 1 ∣ < 2 ) = 3 4 P(|X - 1| < 2) = \frac{3}{4} P ( ∣ X − 1∣ < 2 ) = 4 3 , P ( ∣ X ∣ > 2 ) = 0 P(|X| > 2) = 0 P ( ∣ X ∣ > 2 ) = 0
(b) If X has a uniform distribution in (0, 3) then
V a r ( X ) = ( b − a ) 2 12 = ( 3 − 0 ) 2 12 = 9 12 = 3 4 Var(X) = \frac{{{{\left( {b - a} \right)}^2}}}{{12}} = \frac{{{{\left( {3 - 0} \right)}^2}}}{{12}} = \frac{9}{{12}} = \frac{3}{4} Va r ( X ) = 12 ( b − a ) 2 = 12 ( 3 − 0 ) 2 = 12 9 = 4 3
If Y has exponential distribution with parameter α then
V a r ( Y ) = 1 α 2 Var(Y) = \frac{1}{{{\alpha ^2}}} Va r ( Y ) = α 2 1
V a r ( X ) = V a r ( Y ) ⇒ 3 4 = 1 α 2 ⇒ α 2 = 4 3 ⇒ α = 2 3 3 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} Va r ( X ) = Va r ( Y ) ⇒ 4 3 = α 2 1 ⇒ α 2 = 3 4 ⇒ α = 3 2 3
P ( Y > 0.5 ∣ Y < 1 ) = ∫ 0.5 1 α e − α x d x = ∫ 0.5 1 2 3 3 e − 2 3 3 x d x = − e − 2 3 3 x ∣ 0.5 1 = − e − 2 3 3 + e − 3 3 ≈ 0.246 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 P ( Y > 0.5∣ Y < 1 ) = 0.5 ∫ 1 α e − αx d x = 0.5 ∫ 1 3 2 3 e − 3 2 3 x d x = − e − 3 2 3 x ∣ ∣ 0.5 1 = − e − 3 2 3 + e − 3 3 ≈ 0.246
Answer: α = 2 3 3 \alpha = \frac{{2\sqrt 3 }}{3} α = 3 2 3 , P ( Y > 0.5 ∣ Y < 1 ) ≈ 0.246 P(Y > 0.5|Y < 1) \approx 0.246 P ( Y > 0.5∣ Y < 1 ) ≈ 0.246
Comments