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Given two independent random variables $X\sim\Gamma(s,r)$ and $Y\sim\Gamma(t,u)$, what is the distribution of the difference, i.e. $D=X−Y$? I assume that $s$ and $t$ are integers. How can I obtain the skewness of the difference distribution?

kjetil b halvorsen
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zmgao
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    Duplicate of Difference of Gamma random variables – Dilip Sarwate May 14 '14 at 03:01
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    It's only a partial duplicate because the question and answer in the other post doesn't respond to the skewness question, which can be answered. I don't think this one should close. – Glen_b May 14 '14 at 04:53
  • @Glen_b: Thank you for your help. I really need the closed form solution of skewness. But it looks like very difficult. – zmgao May 15 '14 at 01:20
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    Um, no - my answer all but gives it to you. You merely need to substitute the gamma parameters into the formulas for the second and third moments of a gamma and then put those into the formula at the bottom of my answer. If that takes you more than two minutes, I'd be amazed. – Glen_b May 15 '14 at 01:34
  • @Glen_b: I also want to obtain the kurtosis. But I don't know how to calculate the u4 from MGF? Would please help me? Thank you! – zmgao May 15 '14 at 13:55
  • Is this for some subject? – Glen_b May 15 '14 at 18:48
  • If so, please read the self-study tag wiki info and add the tag. If not, why must you use the MGF? It's doable, but the integral is easy by comparison. But if you really want the kurtosis, you might consider the cumulant generating function instead. – Glen_b May 15 '14 at 18:54

1 Answers1

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Even if we don't have the distribution in closed form, the skewness we can get somewhere with.

For example:

$\,E(D^3) = E((X-Y)^3)\\ \quad\quad\quad= E(X^3)-3E(X^2Y)+3E(XY^2)-E(Y^3) \\ \quad\quad\quad= E(X^3) - 3E(X^2)E(Y)+3E(X)E(Y^2)-E(Y^3)$

and so on with the lower order moments. As a result, $E[(D-\mu_D)^3]$ and $E[(D-\mu_D)^2]$ can be derived and from that, the skewness of the difference. Alternatively,

$E[(D-\mu_D)^3]\\ \quad\quad=E[(X-Y-\{\mu_X-\mu_Y\})^3]\\ \quad\quad=E[(\{X-\mu_X\}-\{Y-\mu_Y\})^3]$

$\quad\quad=E[(X^*-Y^*)^3]\quad$ where here $^*$ indicates centered variables.

$\quad\quad= E(X^{*3}) - 3E(X^{*2})E(Y^{*})+3E(X^{*})E(Y^{*2})-E(Y^{*3})\quad$ as before

$\quad\quad= E(X^{*3}) - 3\text{Var}(X)\cdot 0+3\cdot 0\cdot\text{Var}(Y) -E(Y^{*3})\quad$

$\quad\quad= E(X^{*3}) -E(Y^{*3})\quad$

Similarly, $\text{Var}(D) = \text{Var}(X)+\text{Var}(Y)$.

Or, even more alternatively, we could have just relied on additivity of cumulants to arrive at the same results.

Consequently, $\gamma_1(D) = \frac{\mu_3(D)}{\mu_2(D)^{3/2}}= \frac{\mu_3(X)-\mu_3(Y)}{(\text{Var}(X)+\text{Var}(Y))^{3/2}}$

(Added additional detail for OP):

And of course, $\mu_3(X)=\gamma_1(X)\sigma_X^3$. So we have the skewness of the difference in terms of the skewness and standard deviation of the original variables - so far this is a general result; it doesn't rely on the variables being gamma random variables.

After that it's simple substitution.

Glen_b
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