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Chapter 2: Problem 8

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

Short Answer

Expert verified
The conditional mean \(E(Y | X = x)\) for the given joint pdf is \(x + 1\).

Step by step solution

01

Find Conditional Distribution

In order to find the conditional mean of \(Y\) given \(X = x\), we first need to find the conditional distribution of \(Y\) given \(X = x\). The conditional pdf \(f(y | x)\) is found by dividing the joint pdf by the marginal pdf of \(X\), as follows: \(f(y | x) = \frac{f(x, y)}{f_X(x)}\). However, in this case, we notice that for \(0 < x < y < \infty\), \(f(x, y) = 2 \exp\{- (x + y)\}\), and \(f_X(x) = 2 \exp\{- x\}\). Therefore, \(f(y | x) = \frac{2 \exp\{- (x + y)\}}{2 \exp\{- x\}} = \exp\{- y\}\) for \(x < y < \infty\), zero elsewhere.
02

Compute Conditional Expectation

The definition of conditional expectation is given by \(E(Y | X = x) = \int_{-\infty}^{\infty} yf(y | x)\,dy\). Thus, we need to compute the integral \(\int_{x}^{\infty} y\exp\{- y\}dy\). This is an improper integral, but we can solve it by integrating by parts, where \(u=y\), \(dv=\exp\{- y\}dy\), then \(du=dy\), and \(v=-\exp\{- y\}\). Applying the formula for integration by parts, we find that \(\int_{x}^{\infty} y\exp\{- y\}dy = [-y \exp\{- y\}]_{x}^{\infty} - \int_{x}^{\infty} -\exp\{- y\}dy = x \exp\{- x\} + \exp\{- x\}\).
03

Simplify Solution

The solution of Step 2 simplifies to \(E(Y | X = x) = x + 1\), which is the conditional expectation of \(Y\) given \(X = x\).

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Most popular questions from this chapter

Let 13 cards be talsen, at random and without replacement, from an ordinary deck of playing cards. If \(X\) is the number of spades in these 13 cards, find the pmf of \(X\). If, in addition, \(Y\) is the number of hearts in these 13 cards, find the probability \(P(X=2, Y=5) .\) What is the joint pmf of \(X\) and \(Y ?\)

Let \(X\) and \(Y\) be random variables with the space consisting of the four points: \((0,0),(1,1),(1,0),(1,-1)\). Assign positive probabilities to these four points so that the correlation coefficient is equal to zero. Are \(X\) and \(Y\) independent?

If the correlation coefficient \(\rho\) of \(X\) and \(Y\) exists, show that \(-1 \leq \rho \leq 1\). Hint: Consider the discriminant of the nonnegative quadratic function $$h(v)=E\left\\{\left[\left(X-\mu_{1}\right)+v\left(Y-\mu_{2}\right)\right]^{2}\right\\}$$ where \(v\) is real and is not a function of \(X\) nor of \(Y\).

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 \(X_{1}, X_{2}, X_{3}\) be iid with common mgf \(M(t)=\left((3 / 4)+(1 / 4) e^{t}\right)^{2}\), for all \(t \in R\) (a) Determine the probabilities, \(P\left(X_{1}=k\right), k=0,1,2\). (b) Find the mgf of \(Y=X_{1}+X_{2}+X_{3}\) and then determine the probabilities, \(P(Y=k), k=0,1,2, \ldots, 6\)
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