The skewness of the difference between two independent random variables, X and Y, is given by:
$$\text{skewness}(X-Y) = \frac{\mu _3(X) - \mu _3(Y)}{\big(\mu _2(X) + \mu _2(Y)\big)^{3/2}}$$
But what is the expression for kurtosis(X - Y)?
The skewness of the difference between two independent random variables, X and Y, is given by:
$$\text{skewness}(X-Y) = \frac{\mu _3(X) - \mu _3(Y)}{\big(\mu _2(X) + \mu _2(Y)\big)^{3/2}}$$
But what is the expression for kurtosis(X - Y)?
If we write $Z=X-E[X]$, $W=Y-E[Y]$, then clearly $\mu_4(X-Y)=\mu_4(Z-W)$, while $E[Z]=E[W]=0$ and so $E[Z^n]=\mu_n$ and likewise for $W$. Then $$ \mu_4(Z-W) = E[(Z-W)^4] = E[Z^4]-4E[Z^3]E[W]+6E[Z^2]E[W^2]-4E[Z]E[W^3]+E[Z^4] \\ = E[Z^4]+6E[Z^2]E[W^2]+E[Z^4] \\ = \mu_4(Z) + \mu_4(W) + 6\mu_2(Z)\mu_2(W), $$ by independence and linearity of expectation. Hence, $$ \operatorname{kurt}{(X-Y)} = \operatorname{kurt}{(Z-W)} = \frac{\mu_4(Z) + \mu_4(W) + 6\mu_2(Z)\mu_2(W)}{(\mu_2(Z)+\mu_2(W))^2} \\ = \frac{\mu_4(X) + \mu_4(Y) + 6\mu_2(X)\mu_2(Y)}{(\mu_2(X)+\mu_2(Y))^2} $$
But does it follow it has no definite expression?
– user120911 May 05 '17 at 19:20