DE LAS COMISIONES Y LOS COMITÉS CAPÍTULO I

As we have seen, the density function for a sum of independent random variables is a convolution of densities. This leads to some interesting relations, which are discussed here.

5.6.1 Bernoulli random variables

Suppose Y in (5.48) is a sum of IID Bernoulli random variables. That is, each Xi is

discrete and takes on only values of 1 or 0 with probability p and q=1–p respectively. The probability generating function (PGF) for the Bernoulli random variable reduces to

The PGF of Y is then given by

This PGF is the nth _{power of a binomial, which can be expanded as}

8 _{A similar convolution for discrete random variables can be derived using the PGF.}

The term in square brackets is thus fY[k], which is the binomial PMF (see (3.5)).

It is really not surprising that Y turns out to be a binomial random variable. Since the

Xi are either 1 or 0, Y simply counts the number of 1’s in the sum. You have already seen

when we introduced the binomial PMF that this distribution can be used to specify the probability of k 1’s in a sequence of n binary digits.

5.6.2 Geometric random variables

Let K represent the sum of n IID geometric random variables of Type I (Equation(3.6)). Then the PMF for K can be found (using the PGF) to be

(5.56)

This is known as the Pascal PMF.

If the geometric PMF describes the number of bits to the first error in a stream of binary data, then the Pascal PMF describes the number of bits to the nth _{error in the}

stream. More generally, when observing discrete events, fK[k] is the probability that the nth _{event of interest occurs at the kth event. Some authors refer to this distribution or a}

variation of it as the negative binomial. However, while there is no dispute about the definition of the Pascal PMF, there does not seem to be precise agreement about the definition of the negative binomial [1, 4, 5, 6, 7].

Example 5.9: Digital data is transmitted one byte at a time over a computer interface. Errors in each byte occur and are detected with probability p. If six errors occur within a sequence of ten bytes, a retransmission is requested as soon as the sixth error is detected. What is the probability that a retransmission is requested?

In this example n=6 and the probability of a retransmission is given by

Suppose that on the average, errors occur at the rate of 1 error in 100 bytes. Then by taking p=1/100, the above expression evaluates to Pr=2.029×10–10. This corresponds to

one retransmission in about 5 megabytes of data. □

5.6.3 Exponential random variables

When Y is the sum of n IID exponential random variables, the PDF of Y is given by (5.57)

This is known as the Erlang density function. To interpret this PDF, recall the use and interpretation of the exponential density function. In waiting for events that could occur at any time on a continuous time axis with average arrival rate λ, the exponential density characterizes the waiting time to the next such event. A sum of n exponential random variables Xi can therefore be interpreted as the total waiting time to the nth event. The

Erlang density characterizes this total waiting time.

The Erlang PDF is most easily derived by starting with the MGF for the exponential, namely

raising it to the nth _{power, and inverting. Further use of this density is made in Chapter 8}

in the section on the Poisson process.

The integral of the density (5.57) is also a useful formula. In particular, the cumulative density function is given by

(5.58) The Erlang is a special case of the Gamma PDF which is given by

(5.59)

where the parameter a is any positive real number and the Gamma function Γ(a) is defined by

(5.60) When a is a positive integer n, this integral reduces to Γ(n)=(n–1)!, so the Gamma PDF reduces to the Erlang.

Example 5.10: Consider a supply of four light bulbs. The lifetime of each bulb is an exponential random variable with a mean of 1/λ=2 months. One light bulb is placed in service; when a bulb burns out it is immediately discarded and replaced with another. Let us find the probability that these four light bulbs last more than one year.

Let Xi represent the lifetime of the ith bulb. Then the total lifetime Y is given by

which is an Erlang random variable.

The probability that the total lifetime (in months) is more than 12 is given by

where the last part of the equation comes from using (5.58) with n=4, λ=1/2, and

y=12.

The astute reader may notice the sum on the right side looks like a Poisson PMF, and indeed it is. You can think of “burnouts” as arrival of events. Then the probability of less than 4 burnouts in a period of 12 months is given by where fK is the Poisson

PMF given by (3.8) of Chapter 3. □

5.6.4 Gaussian random variables

In the case where Y is the sum of independent Gaussian random variables (not necessarily identically distributed), Y is also a Gaussian random variable. This is an important fact to

remember. The proof of this result using the MGF is explored in Prob. 5.30 at the end of

this chapter. If the random variables Xi have means mi and variances then Y has a

Gaussian PDF with mean and variance

5.6.5 Squared Gaussian random variables

A final slightly different sum of random variables is

(5.61) where the Xi are zero-mean unit-variance Gaussian random variables (i.e., mi=0, = 1).

(5.62)

where the Gamma function is defined by (5.60). This PDF is known as a Chi-square density function with “n degrees of freedom.” This density is best known for its use in the “Chi-squared test,” a statistical procedure used to test for Gaussian random variables. The Chi-squared density is also a special case of the Gamma density function given above.