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Similarly, as a function of x, fX Y Z x y, z is an N y, 1 density. The straightforward approach is to put f c: Hence, Nt is continuous in mean square.

The corresponding confidence interval is [ Here is an example. In other words, when integrated with respect to x, the result is one. Next, define the scalar Y: Hence, by Prob- lem 55 c in Chapter 4 and the remark following it, 2 Z 2 is chi-squared with two degrees of freedom.

Let Ti denote the time to transmit packet i. Let Xi denote the flow on link i, and put Yi: Let A denote the event that Anne catches no fish, and let B denote the event that Betty catches no fish. For any constants c1. Let X1X2X3 be the random digits of the drawing.

### Errata for Probability and Random Processes for Electrical and Computer Engineers

For arbitrary events Fnlet An be as in the preceding problem. Now the event that you test a defective chip is D: To find the Chernoff bound, we must minimize h s: Since the mean is zero, the second moment is also the variance.

This is a causal impulse response. The plan is to show that the increments are Gaussian and uncorrelated. The event that the friend takes two chips is then T: If we multiply both formulass by IB im. Z Since Z is the sum of i. Denote the arrival times of Nt by T1T2.

For the probability measure we take P A: Chapter 3 Problem Solutions 35 Two apples and three carrots corresponds to 0, 0, 1, 1, 0, 0, guvner. Chapter 11 Problem Solutions Otherwise, there is at least one element S of B in A, say ak.

More specifically, there are 96 possibilities for the first packet, 95 for the hubner. We show this to be the case. If neither is zero and one of them is infinity, then the right-hand side is infinity and the inequality is trivial.

Then differentiate to obtain the density.

## Frame ALERT!

These are four disjoint events. Now, since each of the above terms gkbner the third line is a scalar, each term is equal to its transpose. The table inside the back cover of the text gives the nth moment of a gamma random variable. It remains to find the mean and covariance of Y. We now need the following implications: In this case, the integral of the density over the region D must be one.

Hence, they are uncorrelated. Chapter 1 Problem Solutions 7 This is an instance of Problem It is obvious that the Xi are zero mean. It suffices to show that Yn is Cauchy in L p.

Since Xn and Yn each converge in distribution to constants x and soltuions, respectively, they also converge in probability. Third, for disjoint We also note from the text that the sum of two independent Poisson random variables is a Poisson random variable whose parameter is the sum of the individual parameters.

Since the Xi solutiohs i. Recall that in the problem statement, it was shown that every open set U can be written as a countable union of open intervals. Let C be the ball of radius r, C: The true probability is 0.