Chapter 5: Q. 5.14 (page 218)

Suppose that the cumulative distribution function of the random variable $X$ is given by
$F (x) = 1 - e^{- x^{2}} x > 0$
Evaluate $(a) P [X > 2]; (b) P [1 < X < 3)$ ; (c) the hazard rate function of $F_{i} (d) E [X]; (e) Var (X)$ .
Hint: For parts $(d)$ and $(e)$ , you might want to make use of the results of Theoretical Exercise $5.5$ .

Short Answer

Expert verified

(a) The cumulative distribution function of $P (X > 2) = e^{- 4}$

(b) The cumulative distribution function of random variables is $P (1 < X < 3) = e^{- 1} - e^{- 9}$

(d) $E X = \frac{\sqrt{π}}{2}$

(e) $Var (X) = \frac{3 π}{4}$

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

The cumulative distribution of the function $P [X > 2]$

$P (X > 2) = 1 - P (X \leq 2) = 1 - F_{X} (2) = 1 - (1 - e^{- 2^{2}}) P (X > 2) = e^{- 4}$

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

$P (1 < X < 3) = P (X < 3) - P (X < 1) = F_{X} (3) - F_{X} (1) = e^{- 1} - e^{- 9} P (1 < X < 3) = e^{- 1} e^{- 9}$

Step :3 The hazard rate of function (part c)

We have that,

$f_{X} (x) = \frac{d}{d x} F_{X} (x) = \frac{d}{d x} (1 - e^{- x^{2}}) = 2 x e^{- x^{2}} F_{X} (X) = 2 x e^{- x^{2}}$

Here the hazard function

$μ_{X} (x) = \frac{f_{X} (x)}{1 - F_{X} (x)} = \frac{2 x e^{- x^{2}}}{1 - (1 - e^{- x^{2}})} μ_{X} (X) = 2 x$

Step :4 Value of E[X] (part d)

$E X = \int_{ℝ} x f_{X} (x) d x = \int_{0}^{\infty} x \cdot 2 x e^{- x^{2}} d x = \int_{0}^{\infty} 2 x^{2} e^{- x^{2}} d x$

$u = x and d v = 2 x e^{- x^{2}} d x . Thus d u = d x and v = - e^{- x^{2}}$

$\int_{0}^{\infty} 2 x^{2} e^{- x^{2}} d x = - {x \cdot e^{- x^{2}}|}_{0}^{\infty} + \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2}$

Step : 5 variable Var⁡(X)

In order to calculate the variance, let's find the second moment. We ve got

$E (X^{2}) = \int_{0}^{\infty} x^{2} f_{X} (x) d x = \int_{0}^{\infty} 2 x^{3} e^{- x^{2}} d x$

$u = x^{2} and d v = 2 x e^{- x^{2}} d x . Thus d u = 2 x d x and v = - e^{- x^{2}}$

$\int_{0}^{\infty} 2 x^{3} e^{- x^{2}} d x = - {x^{2} \cdot e^{- x^{2}}|}_{0}^{\infty} + \int_{0}^{\infty} 2 x e^{- x^{2}} d x = {e^{- x^{2}}|}_{0}^{\infty} = 1$

Hence, the variance is adequate to

$Var (X) = E (X^{2}) - E (X)^{2} = 1 - \frac{π}{4} = \frac{3 π}{4}$

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Short Answer

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

Step :3 The hazard rate of function (part c)

Step :4 Value of E[X] (part d)

Step : 5 variable Var⁡(X)

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Suppose that the cumulative distribution function of the random variable Xis given byF(x)=1−e−x2x>0Evaluate (a)P[X>2];(b)P[1<X<3); (c) the hazard rate function of Fi(d)E[X]; (e)Var⁡(X).Hint: For parts (d)and (e), you might want to make use of the results of Theoretical Exercise 5.5.

Short Answer

Step by step solution

Step :1 The cumulative distribution of the function P[X>2] (part a)

Step :2 The cumulative distribution of the function P[1<X<3) (part b)

Step :3 The hazard rate of function (part c)

Step :4 Value of E[X] (part d)

Step : 5 variable Var⁡(X)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

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Calculus

Study anywhere. Anytime. Across all devices.