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

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 $$X_1-X_2$$?

• 1. Are $X_1$ and $X_2$ intended to be independent? 2. "I am not interested in the absolute difference as it collapses to another binomial distribution" no, in general it doesn't. The absolute difference of two sums is not the same thing as the sum of absolute differences. 3. Presumably you seek the moment-skewness ($\mu_3/\sigma^{3})$, rather than some other skewness measure. Is that correct? Nov 4, 2022 at 0:55
• 4. Your title says "binomial averages" but it looks like $X_1$ and $X_2$ are sums of Bernoullis, not averages, so $X_1-X_2$ is simply the difference of two binomial variates. You might say "binomial random variables" or "binomial sums" or even just "binomials" in the title but it doesn't look like the "averages" fits here. Nov 4, 2022 at 1:26

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
$$\sigma_1^2+\sigma_2^2= n_1p_1(1-p_1)+n_2p_2(1-p_2)$$
$$\frac{n_1p_1(1-p_1) (1-2p_1) -n_2p_2(1-p_2) (1-2p_2)}{[n_1p_1(1-p_1)+n_2p_2(1-p_2)]^{\frac{3}{2}}}$$