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Chapter 2: Problem 8

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=3 x, 0

Short Answer

Expert verified
The random variables X and Y are not independent and the conditional expectation of X given Y is \(E[X|Y] = 2/3(1 - y^3)/(1 - y^2)\), for 0<y<1.

Step by step solution

01

Verifying Independence

Firstly, check if the joint PDF factors into a product of individual PDFs. If it does, then \(X\) and \(Y\) are independent, otherwise they are not. Here, \(f(x, y) = 3x\) in the region 0<y<x<1. This joint PDF can't be factored into a product of individual PDFs, hence X and Y are not independent.
02

Compute the Conditional PDF

Since \(X\) and \(Y\) are not independent, the conditional PDF of \(X\) given \(Y\) should be calculated. We can do so by dividing the joint PDF by the marginal PDF of \(Y\). However, the marginal PDF of \(Y\) can be obtained by integrating the joint PDF over all values of \(X\), given \(Y\). In this case, the marginal PDF of Y, \(f_Y(y)\), can be obtained by integrating from y to 1 (the limits of x when y is given), i.e., \(f_Y(y)\) = \(\int_{y}^{1}f(x, y) dx = \int_{y}^{1} 3x dx = 3(\frac{x^2}{2})\Big|_{y}^{1} = 3(\frac{1}{2}- \frac{y^2}{2}) = 3/2 - 3y^2/2\).
03

Calculate the Conditional PDF

Next, we compute the conditional PDF of X given Y, represented as \(f_{X|Y}(x|y)\). This is computed as the ratio of the joint PDF to the marginal PDF of Y. Therefore, \(f_{X|Y}(x|y)\) = \( f(x, y)/f_Y(y) = 3x/(3/2 - 3y^2/2) = 2x / (1 - y^2)\). This is valid for y < x < 1
04

Compute Conditional Expectation

Now, we calculate the conditional expectation, \(E(X|y)\). This is evaluated as the integral of \(x * f_{X|Y}(x|y)\) over the range of \(X\) given \(Y\). Hence, \(E(X|y) = \int_{y}^{1} x*f_{X|Y}(x|y) dx = \int_{y}^{1} 2x^2 / (1 - y^2) dx = 2(\frac{x^3}{3})\Big|_{y}^{1} /(1-y^2) = 2/3(1 - y^3)/(1 - y^2), for 0<y<1.

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Most popular questions from this chapter

II the random variables \(X_{1}\) and \(X_{2}\) have the joint \(\operatorname{pd} \iint\left(x_{1}, x_{2}\right)=2 e^{-x_{1}-x_{2}}, 0<\) \(x_{1}

Let \(\mu\) and \(\sigma^{2}\) denote the mean and variance of the random variable \(X\). Let \(Y=c+b X\), where \(b\) and \(c\) are real constants. Show that the mean and variance of \(Y\) are, respectively, \(c+b \mu\) and \(b^{2} \sigma^{2}\).

Suppose that a man leaves for work between 8:00 a.m. and \(8: 30\) a.m. and takes between 40 and 50 minutes to get to the office. Let \(X\) denote the time of departure and let \(Y\) denote the time of travel. If we assume that these random variables are independent and uniformly distributed, find the probability that he arrives at the office before \(9: 00\) a.m.

Let \(f(x)\) and \(F(x)\) denote, respectively, the pdf and the cdf of the random variable \(X\). The conditional pdf of \(X\), given \(X>x_{0}, x_{0}\) a fixed number, is defined by \(f\left(x \mid X>x_{0}\right)=f(x) /\left[1-F\left(x_{0}\right)\right], x_{0}x_{0}\right)\) is a pdf. (b) Let \(f(x)=e^{-x}, 02 \mid X>1)\).

Let \(p\left(x_{1}, x_{2}\right)=\frac{1}{16}, x_{1}=1,2,3,4\), and \(x_{2}=1,2,3,4\), zero elsewhere, be the joint pmf of \(X_{1}\) and \(X_{2} .\) Show that \(X_{1}\) and \(X_{2}\) are independent.
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