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Chapter 2: Problem 7

Let \(X\) and \(Y\) have the pdf \(f(x, y)=1,0

Short Answer

Expert verified
The cumulative distribution function (CDF) of \(Z\) is \(F(z) = 1 - z\ln z\) for \(0<z<1\) and \(F(z) = 1\) for \(z \geq 1\). The probability density function (PDF) of \(Z\) is \(f(z) = -\ln z - 1\) for \(0<z<1\) and \(f(z) = 0\) for \(z \geq 1\).

Step by step solution

01

Understanding the Joint Distribution

The joint pdf of \(X\) and \(Y\) is given as \(f(x, y) = 1\) for \(0<x<1,0<y<1\), which indicates that \(X\) and \(Y\) are independently and uniformly distributed between 0 and 1.
02

Finding the CDF of Z

To find the cumulative distribution function (CDF) of \(Z\), determine the probability that \(Z \leq z\). Since \(Z = X*Y\), we want to find \(P(X*Y \leq z)\). For \(0< z < 1\), the cdf of \(Z\) will be given by \(F(z) = \int_0^1\int_0^\frac{z}{y} dy dx = \int_0^1 (1 - \frac{z}{x})dx = 1 - z\ln z\) for \(0<z<1\). For \(z \geq 1\), \(F(z) = 1\), which indicates the maximum possible value of the cdf.
03

Finding the PDF of Z

To find the probability density function (PDF) of \(Z\), differentiate the CDF of \(Z\) with respect to \(z\). Thus, \(f(z) = \frac{d}{dz}F(z) = -\ln z - 1\) for \(0<z<1\) and \(f(z) = 0\) for \(z \geq 1\).

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Most popular questions from this chapter

Let \(X\) and \(Y\) have the joint pmf described as follows: $$\begin{array}{c|cccccc}(x, y) & (1,1) & (1,2) & (1,3) & (2,1) & (2,2) & (2,3) \\ \hline p(x, y) & \frac{2}{15} & \frac{4}{15} & \frac{3}{15} & \frac{1}{15} & \frac{1}{15} & \frac{4}{15} \end{array}$$ and \(p(x, y)\) is equal to zero elsewhere. (a) Find the means \(\mu_{1}\) and \(\mu_{2}\), the variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), and the correlation coefficient \(\rho\). (b) Compute \(E(Y \mid X=1), E\left(Y \mid X=2\right.\) ), and the line \(\mu_{2}+\rho\left(\sigma_{2} / \sigma_{1}\right)\left(x-\mu_{1}\right) .\) Do the points \([k, E(Y \mid X=k)], k=1,2\), lie on this line?

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=6(1-x-y), x+y<1,0

Find the probability of the union of the events \(a

Five cards are drawn at random and without replacement from an ordinary deck of cards. Let \(X_{1}\) and \(X_{2}\) denote, respectively, the number of spades and the number of hearts that appear in the five cards. (a) Determine the joint pmf of \(X_{1}\) and \(X_{2}\). (b) Find the two marginal pmfs. (c) What is the conditional pmf of \(X_{2}\), given \(X_{1}=x_{1} ?\)

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=2 \exp \\{-(x+y)\\}, 0
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