# Behavior of Gamma Distribution over time

I need to prove that the age behavior of the Gamma Distribution with probability density function $$f(x)=\frac{\lambda^{\alpha}x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)}$$
For $x\geq 0$, $\lambda, \alpha >0$; I.e., the conditional pprobability $P(X>x+t|X>t)$, increases in $t$ whenever $\alpha >1$ and decreases in $t$ whenever $0<\alpha <1$.

Thus far, I have the following: $$P(X>x+t|X>t)=\frac{P(X>x+t)\cap P(X>t)}{P(X>t)} = \frac{P(X>x+t)}{P(X>t)} =\frac{\int_{x+t}^{\infty}\frac{\lambda^{\alpha}u^{\alpha-1}e^{-\lambda u}}{\Gamma(\alpha)}du}{\int_{t}^{\infty}\frac{\lambda^{\alpha}u^{\alpha-1}e^{-\lambda u}}{\Gamma(\alpha)}du}.$$

I didn’t know how to go further, until my professor hinted to take the derivative of the last part with respect to $\alpha$. If it had been w.r.t. $t$, then I could use the fundamental theorem of calculus maybe to help me figure out the derivatives of those integrals.

I do know that $\Gamma(\alpha)=\int_{0}^{\infty}y^{\alpha – 1}e^{-y}dy = (\alpha-1)\Gamma(\alpha -1)$, but how do I take the derivative of that w.r.t. $\alpha$? At least in a meaningful way that I should help me towards my ultimate goal, which is to prove that the conditional probability increases in $t$ for $\alpha >1$ and decreases in $t$ for $0<\alpha<1$.

I’ve even tried even entering this all into Maple and letting it do the calculations for me; the problem with this is I keep getting error messages because I don’t know how to define $\alpha$ properly! I’m so lost and desperately need to see how these derivatives are done. Even if you don’t want to work out the whole thing for me, I’d settle just to see how to find the derivative of the integral in the numerator of my last expression. I would accept an answer with just this part worked out in detail. Please.
Fix some $a>0$ and consider, for every $x>0$, $$g(x)=x^{a-1}e^{-x}\qquad G(x)=\int_x^\infty g(y)dy$$ Then the question asks to study the sense of variation of the function $R_x$ defined on $t>0$ as $$R_x(t)=\frac{G(x+t)}{G(t)}$$ Equivalently, considering the logarithmic derivative of $R_x$ and using the fact that $G’=-g$, one is looking for the sign of $$h(t,x)=g(t)G(x+t)-g(t+x)G(t)$$
The change of variable $y\to zy$ in $G(z)$ yields, for every $z>0$, $$G(z)=z\int_1^\infty g(zy)dy=z^a\int_1^\infty y^{a-1}e^{-zy}dy$$ hence, applying this to $z=t$ and to $z=x+t$, one sees that $$h(t,x)=t^{a-1}e^{-t}(x+t)^a\int_1^\infty y^{a-1}e^{-(x+t)y}dy-(t+x)^{a-1}e^{-t-x}t^a\int_1^\infty y^{a-1}e^{-ty}dy$$ which has the sign of $$j(t,x)=(x+t)\int_1^\infty y^{a-1}e^{-(x+t)y}dy-e^{-x}t\int_1^\infty y^{a-1}e^{-ty}dy$$ Now, integrating by parts, for every $z>0$, $$z\int_1^\infty y^{a-1}e^{-zy}dy=e^{-z}+(a-1)\int_1^\infty y^{a-2}e^{-zy}dy$$ hence $$j(t,x)=e^{-x-t}+(a-1)\int_1^\infty y^{a-2}e^{-(x+t)y}dy-e^{-x}\left(e^{-t}+(a-1)\int_1^\infty y^{a-2}e^{-ty}dy\right)$$ which has the sign of $(a-1)k(t,x)$ with $$k(t,x)=\int_1^\infty y^{a-2}e^{-(x+t)y}dy-e^{-x}\int_1^\infty y^{a-2}e^{-ty}dy$$ This is also $$k(t,x)=\int_1^\infty y^{a-2}(e^{-xy}-e^{-x})e^{-ty}dy$$ The parenthesis in the integral is always negative hence $k(t,x)<0$ thus $R_x(t)$ is an increasing function of $t$ if $a<1$ and a decreasing function of $t$ if $a>1$.
$$\frac{\partial R_x(t)}{\partial t}=(1-a)\frac{g(t)g(x+t)}{G(t)^2}\int_0^\infty e^{-xz}\int_z^\infty(y+1)^{a-2}e^{-ty}dy\,dz$$