Question

Let X have the binomial distribution with parameters n and p. Let Y have the binomial distribution with parameters n and p/k with k > 1. Let \(Z = kY\).

a. Show that X and Z have the same mean.

b. Find the variances of X and Z. Show that, if p is small, then the variance of Z is approximately k times as large as the variance of X.

c. Show why the results above explain the higher variability in the bar heights in Fig. 6.2 compared to Fig. 6.1.

Short answer

a. It is showed that X and Z have the same mean.

b. Variance of X is \(np\left( {1 - p} \right)\) and variance ofZ is \(knp\left( {1 - \frac{p}{k}} \right)\) .Also, It is proved that if p is small, then the variance of Z is approximately k times as large as the variance of X.

c. By part (b), one expect the heights in Fig.6.2 to have approximately twice the variance of the heights in Fig.6.1.

Step-by-step solution

Given information

Let X follows binomial distribution with parameter n and p. Also, Y follows binomial distribution with parameters n and p/k with k>1.

Calculating mean of X and Z

Let, \(Z = kY\)

Since, Y follows binomial distribution with parameters n and p/k with k>1.

Therefore,

\(E\left( Y \right) = n.\frac{p}{k}\)

Now,

\(\begin{array}{c}E\left( Z \right) = kE\left( Y \right)\\ = k.n.\frac{p}{k}\\ = np\end{array}\)

But, \(E\left( X \right) = np\)

Hence, X and Z have the same mean.

Calculating variance of X and Z

Since, X follows binomial distribution with parameter n and p.

Therefore,

\(V\left( X \right) = np\left( {1 - p} \right)\)

Also,

Y follows binomial distribution with parameter n and p/k.

Therefore,

\(V\left( Y \right) = n\frac{p}{k}\left( {1 - \frac{p}{k}} \right)\)

So,

\(\begin{array}{c}V\left( {kY} \right) = {k^2}V\left( Y \right)\\ = {k^2}n\frac{p}{k}\left( {1 - \frac{p}{k}} \right)\\ = knp\left( {1 - \frac{p}{k}} \right)\end{array}\)

If p is small, then both \(\left( {1 - p} \right)\) and \(\left( {1 - \frac{p}{k}} \right)\) will be close to 1, and \(V\left( Z \right)\) is approximately knpwhile the variance of X is approximately np.

Explanation for the above results

In Fig. 6.1, each bar has height equal to 0.01 times a binomial random variable with parameters 100 and the probability that \({X_1}\) is in the interval under the bar.

In Fig. 6.2, each bar has height equal to 0.02 times a binomial random variable with parameters 100 and probability that\({X_1}\) is in the interval under the bar.

The bars in Fig.6.2 have approximately one-half of the probability of the bars in Fig.6.1, but their heights have been multiplied by 2. By part (b), we expect the heights in Fig.6.2 to have approximately twice the variance of the heights in Fig.6.1.