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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 $0<\alpha<1$.

Could somebody please help me figure out how to finish this proof? Thank you for your time and patience.

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.

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  • $\begingroup$ If $f$ is log-convex (log-concave), then for any $x\ge 0$ $f(x+t)/f(t)$ increases (decreases). So differentiate $\log P(X>t)$ twice and check whether you can prove that the result is positive or negative. $\endgroup$ – zhoraster Sep 20 '17 at 5:55
  • $\begingroup$ @zhotaster I've never even heard of log-convex before. I asked my Prof after class last night and he said to differentiate the quotient of those two integrals, I think with respect to t. But I'm not even sure how to do that, since t doesn't really appear in any of them. Could you help me with that? $\endgroup$ – user100463 Sep 20 '17 at 17:40
  • $\begingroup$ @zhoraster I've updated the question to better reflect what assistance I currently need. $\endgroup$ – user100463 Sep 20 '17 at 18:27
  • $\begingroup$ @zhoraster anyway, why would I just differentiate $\log P(X>t)$? Why not $\log((P(X>x+t))/(P(X>t)))$? Anyway, in order to differentiate just $\log P(X>t)$, I need to be able to differentiate $P(X>t)$ (with respect to what, by the way?), but I have tried to differentiate it, Gamma functions keep popping up. I tried putting it into a computer algebra system - the result it gave me was horrible. Unless I'm doing something wrong. You seem to be pretty sure this will work. Can you please work out the details in finding that derivative for me? $\endgroup$ – user100463 Sep 20 '17 at 20:20
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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 $a>1$.


Edit: To sum up the sequence of computations above, $$\frac{\partial R_x(t)}{\partial t}=(1-a)\frac{g(t)g(x+t)}{G(t)^2}\int_0^\infty e^{-xz}\int_z^\infty(y+1)^{a-2}e^{-ty}dy\,dz$$

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  • $\begingroup$ But taking a derivative with respect to $\alpha$ isn't a bad idea! I would try doing it. $\endgroup$ – zhoraster Sep 21 '17 at 5:12
  • $\begingroup$ @zhoraster "that [taking a derivative with respect to α] is the question you should think about." Who, me? Actually, differentiating everything with respect to $\alpha$ introduces logarithms and I see little hope that this could lead to the desired result. And anyway, we do have a proof, right? $\endgroup$ – Did Sep 21 '17 at 9:46
  • $\begingroup$ Not you, but ALannister. He deleted his comments, so now mine look a bit out of place. Personally, I find your answer clear enough (certainly clearer than my average answer). $\endgroup$ – zhoraster Sep 21 '17 at 12:02
  • $\begingroup$ @zhoraster Understood. Sorry for the noise. $\endgroup$ – Did Sep 21 '17 at 12:34
  • $\begingroup$ @Did there were some difficulties I was having, like how the integral becomes $\int_{1}^{\infty}$ as opposed to $\int_{z}^{\infty}$ when you did the change of variables $y \mapsto yz$. I think it's because of that very same change of variables; originally, we had limits of integration that were $y=x$-values, then those $y$- values became $z=yz$- values, and because $z>0$, we can divide through to get $y=1$. Is that your reasoning? $\endgroup$ – user100463 Sep 21 '17 at 14:14

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