Question

Suppose \(X\) is \(b(n, p)\). Then by definition the pmf is symmetric if and only if \(p(x)=p(n-x)\), for \(x=0, \ldots, n\). Show that the pmf is symmetric if and only if \(p=1 / 2\)

Short answer

The pmf of a binomial distribution is symmetric if and only if \( p = 1/2 \)

Step-by-step solution

Formula for pmf under binomial distribution

The pmf for a binomial distribution is given by: \( p(x; n, p) = C(n, x) p^x (1-p)^{n-x} \) where \(C(n, x)\) is the binomial coefficient given by \( C(n, x) = n! / [x!(n-x)!] \)

Equation for p(x) and p(n-x)

The function \(p(x)\) is equal to \(C(n, x) p^x (1-p)^{n-x}\) and \(p(n-x)\) is equal to \( C(n, n-x) p^{n-x} (1-p)^x \). Since \(C(n, x) = C(n, n-x)\), we can say that \(p(x)=p(n-x)\) if and only if \(p^x (1-p)^{n-x} = p^{n-x} (1-p)^x \)

Mathematical Manipulation

Following from the previous steps, for \(p(x) = p(n-x)\), we must have that \(p^x (1-p)^{n-x} = p^{n-x} (1-p)^x \). Simplifying this, the equation implies \( p / (1-p) = (n - 2x) / x \)

Existence of Unique solution

Solving the equation \( p / (1-p) = (n - 2x) / x \) for p, we get \( p = (n - 2x) / (n - x) \). The only value that would make the pmf symmetric for all x is \( p = 1/2 \)