/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The random variables \(X\) and \... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 23

The random variables \(X\) and \(Y\) have joint density function. $$ f(x, y)=12 x y(1-x) \quad 0

Short Answer

Expert verified
(a) X and Y are independent. (b) \(E[X] = \frac{1}{2}\). (c) \(E[Y] = \frac{2}{3}\). (d) \(\operatorname{Var}(X) = \frac{7}{20}\). (e) \(\operatorname{Var}(Y) = \frac{1}{18}\).

Step by step solution

01

(a) Checking for Independence of X and Y

To check for independence, we need to find the marginal density functions \(f_X(x)\) and \(f_Y(y)\) as follows: \(f_X(x) = \int_{0}^{1} f(x,y) dy\) and \(f_Y(y) = \int_{0}^{1} f(x,y) dx\) Computing \(f_X(x)\): \(f_X(x) = \int_{0}^{1} 12 x y (1-x) dy = 12x(1-x) \int_{0}^{1} y dy = 12x(1-x)\frac{1}{2}\) Computing \(f_Y(y)\): \(f_Y(y) = \int_{0}^{1} 12 x y (1-x) dx = y \int_{0}^{1} 12 x (1-x) dx = y \left[6x^2 - 4x^3 \right]_0^1 = 2y\) Now, to check for independence, we can compute the product \(f_X(x) f_Y(y)\) and compare to \(f(x,y)\): \(f_X(x) f_Y(y) = 12x(1-x)\frac{1}{2}(2y) = 12xy(1-x)\) Since \(f(x,y) = f_X(x) f_Y(y)\), we can conclude that X and Y are independent.
02

(b) Finding \(E[X]\)

To find the expected value of X, we will compute \(E[X] = \int xf_X(x)dx\): \(E[X] = \int_{0}^{1} x(12x(1-x)\frac{1}{2}) dx = 6\int_{0}^{1} x^2 (1-x) dx\) \( = 6( \int_{0}^{1} x^2 dx - \int_{0}^{1} x^3 dx ) = 6( \frac{1}{3} - \frac{1}{4} ) = \frac{1}{2}\) So, \(E[X] = \frac{1}{2}\).
03

(c) Finding \(E[Y]\)

To find the expected value of Y, we will compute \(E[Y] = \int yf_Y(y)dy\): \(E[Y] = \int_{0}^{1} y(2y) dy = 2\int_{0}^{1} y^2 dy = 2 \frac{1}{3} = \frac{2}{3}\) So, \(E[Y] = \frac{2}{3}\).
04

(d) Finding \(\operatorname{Var}(X)\)

To find the variance of X, we will compute \(\operatorname{Var}(X) = E[X^2] - E[X]^2\): First, we need to find \(E[X^2]\): \(E[X^2] = \int_{0}^{1} x^2(12x(1-x)\frac{1}{2})dx = 6\int_{0}^{1} x^3(1-x) dx\) \(= 6( \int_{0}^{1} x^3 dx - \int_{0}^{1} x^4 dx ) = 6( \frac{1}{4} - \frac{1}{5} ) = \frac{3}{5}\) Now, computing \(\operatorname{Var}(X)\): \(\operatorname{Var}(X) = E[X^2] - E[X]^2 = \frac{3}{5} - (\frac{1}{2})^2 = \frac{3}{5} - \frac{1}{4} = \frac{7}{20}\) So, \(\operatorname{Var}(X) = \frac{7}{20}\).
05

(e) Finding \(\operatorname{Var}(Y)\)

To find the variance of Y, we will compute \(\operatorname{Var}(Y) = E[Y^2] - E[Y]^2\): First, we need to find \(E[Y^2]\): \(E[Y^2] = \int_{0}^{1} y^2(2y) dy = 2\int_{0}^{1} y^3 dy = 2( \frac{1}{4} ) = \frac{1}{2}\) Now, computing \(\operatorname{Var}(Y)\): \(\operatorname{Var}(Y) = E[Y^2] - E[Y]^2 = \frac{1}{2} - (\frac{2}{3})^2 = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}\) So, \(\operatorname{Var}(Y) = \frac{1}{18}\).

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Most popular questions from this chapter

An ambulance travels back and forth, at a constant speed, along a road of length \(L\). At a certain moment of time an accident occurs at a point uniformly distributed on the road. [That is, its distance from one of the fixed ends of the road is uniformly distributed over \((0, L)\).] Assuming that the ambulance's location at the moment of the accident is also uniformly distributed, compute, assuming independence, the distribution of its distance from the accident.

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Suppose that \(F(x)\) is a cumulative distribution function. Show that (a) \(F^{\prime \prime}(x)\) and (b) \(1-[1-F(x)]^{n}\) are also cumulative distribution functions when \(n\) is a positive integer. HINT: Let \(X_{1}, \ldots, X_{n}\) be independent random variables having the common distribution function \(\vec{F}\). Define random variables \(Y\) and \(Z\) in terms of the \(X\) so that \(P\\{Y \leq x\\}=F^{n}(x)\), and \(P\\{Z \leq x\\}=1-[1-F(x)]^{n}\).

The joint probability mass function of \(X\) and \(Y\) is given by $$ \begin{array}{ll} p(1,1)=\frac{1}{8} & p(1,2)=\frac{1}{4} \\ p(2,1)=\frac{1}{8} & p(2,2)=\frac{1}{2} \end{array} $$ (a) Compute the conditional mass function of \(X\) given \(Y=i, i=1,2\). (b) Are \(X\) and \(Y\) independent? (c) Compute \(P\\{X Y \leq 3\\}, P\\{X+Y>2\\}, P\\{X / Y>1\\}\).
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