Homework 7 Solutions - Statistics Department



Homework 7 Solutions

1.

A Poisson process with [pic] accidents per week is equivalently a Poisson process with [pic] accidents per day. Thus, given that 7 accidents occurred in a particular week, the probability that exactly one accident occurred each day of that week is just

[pic].

For a Poisson process with parameter [pic], the waiting times between arrivals (in this case, accidents) are independent and exponentially distributed with mean [pic]. In this situation, that means the waiting times are exponentially distributed with mean 0.1 weeks between accidents, which corresponds to a mean of 16.8 hours between accidents. However, the distribution of an exponential random variable is heavily right-skewed – i.e., much of the mass of the pdf is concentrated below the mean. (See plot below for this exponential distribution over the interval [0, 1]) Hence, it is far more likely that waiting times less than 16.8 hours occur, and with shorter waiting times, it is difficult to have only 1 accident a day for 7 straight days.

[pic]

2. Ross 5.37 p. 251

a) If X is uniformly distributed over (-1, 1), then we have

[pic]

Hence, the probability that [pic] is

[pic].

b) To find the density function of [pic]:

[pic], where [pic].

Taking the derivative of this with respect to a gives us

[pic], where [pic].

In other words, [pic] is uniformly distributed on (0, 1).

3. Ross 5.41 p. 251

We have that [pic] is uniformly distributed on [pic]. If [pic], then to find the distribution of R, we just find the cdf for R as a function of x and then take the derivative with respect to x to obtain the pdf.

[pic]

4. Ross 6.6 p. 313

There are a total of [pic] different sequences in which the 5 transistors (2 defective) can be drawn from the bin. Each of these sequences is equally likely to occur, so the joint pdf of [pic] and [pic] is just

[pic]

5. Ross 6.11 p. 314

The customers can be viewed as draws from a multinomial distribution where [pic] is the probability of an “ordinary” purchaser, [pic] is the probability of a “plasma” purchaser, and [pic] is the probability of a browser. Hence, the probability that the store owner sells exactly 2 ordinary sets and 1 plasma set on a day when 5 customers enter his store is just

[pic].

6. Ross 6.10 p. 314

The joint pdf of X and Y is [pic], where x and y are both non-negative real numbers.

a) To find the probability that X is less than Y:

[pic]

b) For this part, note that the joint pdf of X and Y can be re-written as [pic]; that is, X and Y are independent. Hence, [pic] can be obtained by integrating [pic] and ignoring any mention of y (if X and Y were not independent, we would have to integrate [pic] over the full range of y and then integrate x over the interval (0, a)):

[pic].

7. Ross 6.14 p. 314

Let [pic] and [pic] denote respectively the locations of the ambulance and the accident at the moment the accident occurs. Then the distance is given by [pic], but this is equivalently the range of the two Uniform(0, L) random variables (i.e., [pic]). Using Equation (6.7), we get as the cdf of [pic]

[pic]

To obtain the pdf, take the derivative of the cdf with respect to a:

[pic].

8. Ross 6.15 p. 314

a) The joint density over the region R must integrate to 1, so we have

[pic],

where [pic] is the area of the region R; hence [pic].

b) Since the region R is a square with sides of length 2, [pic], and [pic] for [pic]. But this can be re-written as [pic], where [pic], and [pic]. In other words, X and Y are independent uniform random variables on the interval (-1, 1).

c) The probability that (X, Y) lies in the circle of radius 1 centered at the origin is just the area of the circle multiplied by 1/4:

[pic].

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