Question

Let\(\left\{ {{p_n}} \right\}_{n = 1}^\infty \)be a sequence of numbers such that\(0 < {p_n} < 1\)for all\(n\). Assume that\(\mathop {\lim }\limits_{n \to \infty } {p_n} = p\)with\(0 < p < 1\). Let\({X_n}\)have the binomial distribution with parameters\(k\)and\({p_n}\)for some positive integer\(k\)Prove that\({X_n}\)converges in distribution to the binomial distribution with parameters k and p.

Short answer

\({X_n}\) Converges in distribution to the binomial distribution with parameters k and p

Step-by-step solution

Given information

Let \(\left\{ {{p_n}} \right\}_{n = 1}^\infty \) be the sequence of numbers such that \(0 < {p_n} < 1\)

Verifying \({X_n}\) Converges in distribution to the binomial distribution.

Let Let Fbe the cdf of \({X_n}\).

We know that

\({\rm P}\left( {{X_n} = m} \right) = \left( \begin{align}{l}k\\m\end{align} \right)p_n^m{\left( {1 - {p_n}} \right)^{k - m}}\) for m=0...k and all n.

Hence

\(\mathop {\lim }\limits_{n \to \infty } \left( \begin{array}{l}k\\m\end{array} \right)p_n^m{\left( {1 - {p_n}} \right)^{k - m}} = \left( \begin{array}{l}k\\m\end{array} \right){p^m}{\left( {1 - p} \right)^{k - m}}\).

Also we know that

\(\begin{align}F\left( {{x_n}} \right) &= \sum\limits_{k = 1}^m {{\rm P}\left( {{X_n} = m} \right)} \\ &= \sum\limits_{k = 1}^m {f\left( k \right)} \end{align}\)

\(\begin{align}F\left( {{x_n}} \right) &= F\left( m \right)\\ &= F\left( x \right)\end{align}\)

Hence we can say \({X_n}\) converges in distribution to the binomial distribution with parameters k and p.