What is the skewness of the difference between two binomial random variables?

Question

I have two binomial random variables:

$$ X_1 \sim \text{Binom}(n_1,p_1), \\[6pt] X_2 \sim \text{Binom}(n_2,p_2). $$

I know that the individual skewnesses are:

$$ \mathbb{Skew}(X_1) = \frac{1 - 2p_1}{\sqrt{n_1 p_1 (1-p_1)}}, \\[6pt] \mathbb{Skew}(X_2) = \frac{1 - 2p_2}{\sqrt{n_2 p_2 (1-p_2)}}. \\[6pt] $$

What is the skewness of the difference $X_1-X_2$?

Answer 1

Assuming you do seek the third-moment skewness and assuming independence, the third central moment of $X_1-X_2$ is

\begin{eqnarray} E[((X_1 - X_2) - (\mu_1-\mu_2))^3 ]&=&E[(X_1 - \mu_1 - (X_2 -\mu_2))^3 ]\\\\ &=& E[(X_1-\mu_1)^3] - 3E[(X_1-\mu_1)^2]E(X_2-\mu_2) \\\\ & & \qquad + 3E(X_1-\mu_1)E[(X_2-\mu_2)^2] - E[(X_2-\mu_2)^3]\\\\ &=& E[(X_1-\mu_1)^3] - 0 + 0 - E[(X_2-\mu_2)^3]\\\\ &=&\sigma_1^3 (1-2p_1)/\sqrt{n_1p_1(1-p_1)} \\\\ & & \qquad-\hspace{1mm}\sigma_2^3 (1-2p_2)/\sqrt{n_2p_2(1-p_2)}\\\\ &=&n_1p_1(1-p_1) (1-2p_1) \hspace{1mm} - \hspace{1mm} n_2p_2(1-p_2) (1-2p_2) \end{eqnarray}

The variance of $X_1-X_2$ is

$$\sigma_1^2+\sigma_2^2= n_1p_1(1-p_1)+n_2p_2(1-p_2)$$

Hence (if I made no errors) the skewness should be

$$\frac{n_1p_1(1-p_1) (1-2p_1) -n_2p_2(1-p_2) (1-2p_2)}{[n_1p_1(1-p_1)+n_2p_2(1-p_2)]^{\frac{3}{2}}}$$