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If X is Erlang$(k_1,\lambda)$ and Y is Erlang$(k_2,\lambda)$, then is X+Y Erlang$(k_1+k_2,\lambda)$? Do X and Y need to be independent?

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Yes, the distributions are presumed independent in the context of that factoid. – anon Jul 9 '11 at 7:28
Can you define or give a reference for the distribution "Erlang"? – Rasmus Jul 9 '11 at 12:45
@Rasmus Wikipedia will do. In summary an erlang$(k,\lambda)$ distribution is a $\Gamma(k,1/\lambda)$. It is sufficient that the erlang variables are independent. I suspect that Jim is asking whether it necessary as well. – deinst Jul 9 '11 at 15:06

If $X$ and $Y$ are independent, that's true. This can be easily seen from the interpretation of an Erlang$(k,\lambda)$ as a sum of $k$ iid exponentials, and also from the characteristic function.

If they are not independent, that's not necesarily true. You can take as a counterexample $X=Y$.

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No. In general if we have a continuous joint distribution with pdf $f_{X,Y}(x,y)$ and marginals $f_X(x)$ and $f_Y(y)$ and the sum $Z=X+Y\ $ has pdf $$f_Z(z)=\int_{-\infty}^\infty f_X(z-y)f_Y(y) dy$$ that is not enough to be able to say that $X$ and $Y$ are independent. To see this choose two points $(x_1,y_1)$ and $(x_2,y_2)$ and a small value $\delta$ and a value of $\epsilon$ small enough so that the following construction works.

Define $g_{X,Y}(x,y)$ as follows: In a square of side $\delta$ centered on $(x_1,y_1)$ set $g_{X,Y}(x,y)=f_{X,Y}(x,y)+\epsilon$, similarly for a square of side $\delta$ centered on $(x_2,y_2)$, for the squares of side $\delta$ centered on $(x_1,y_2)$ and $(x_2,y_1)$ set $g_{X,Y}(x,y)=f_{X,Y}(x,y)-\epsilon$, and set $g_{X,Y}(x,y)=f_{X,Y}(x,y)$ everywhere else. This leaves the marginals unchanged but disturbs the sum. Now choose some $d$ larger than $\delta$ and let $(x_3,y_3) = (x_1+d,y_1-d)$ and $(x_4,y_4) = (x_2+d,y_2-d)$, and let $h_{X,Y}(x,y)=g_{X,Y}(x,y)-\epsilon$ for squares of side $\delta$ centered on $(x_3,y_3)$ and $(x_4,y_4)$ and $h_{X,Y}(x,y)=g_{X,Y}(x,y)+\epsilon$ on the squares centered on $(x_3,y_4)$ and $(x_4,y_3)$ and $h_{X,Y}(x,y)=g_{X,Y}(x,y)$ elsewhere. Again this has the same marginal distributions, but now the distribution of $X+Y$ is the same for $f_{X,Y}$ and $h_{X,Y}$, and $X$ and $Y$ are not independent in $h_{X,Y}$.

There is probably a simpler construction that shows this.

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