Ashlyn Krause

2022-07-16

If $1\le \alpha$ show show that the gamma density has a maximum at $\frac{\alpha -1}{\lambda }$.
So using this form of the gamma density function:
$g\left(t\right)=\frac{{\lambda }^{\alpha }{t}^{\alpha -1}{e}^{-\lambda t}}{{\int }_{0}^{\mathrm{\infty }}{t}^{\alpha -1}{e}^{-t}\phantom{\rule{thinmathspace}{0ex}}dt}$
I would like to maximize this. Now i was thinking of trying to differentiate with respect to t, but that is becoming an issue because if i remember correctly the denominator is not a close form when the bound goes to infiniti

Monastro3n

Expert

Step 1
The denominator is a constant as you are integrating it over t. So it is same as maximizing only the numerator. And $\frac{d\left({t}^{\alpha -1}{e}^{-\alpha t}\right)}{dx}=\left(\alpha -1\right){t}^{\alpha -2}{e}^{-\lambda t}-\lambda {t}^{\alpha -1}{e}^{-\lambda t}$.
Step 2
Equating this to 0 you get $t=0,\left(\alpha -1\right)/\lambda$
Check the double derivative at these points to get the maximum.

Caylee Villegas

Expert

Step 1
You're trying to maximize $t↦{t}^{\alpha -1}{e}^{-\alpha t}$ on the interval $0\le t<\mathrm{\infty }$. That is the same as maximizing the logarithm of that function, since the logarithmic function is increasing, and taking the logarithm makes it a bit simpler:
$\mathrm{log}\left({t}^{\alpha -1}{e}^{-\alpha t}\right)=\left(\alpha -1\right)\mathrm{log}t-\alpha t.$
The derivative of this with respect to t is $\frac{\alpha -1}{t}-\alpha =\alpha \frac{1-\left(1/\alpha \right)-t}{t}.$.
Step 2
Since the denominator is positive, the question now is only when the numerator is positive, negative, and zero. The numerator $1-\left(1/\alpha \right)-t$ is positive if $t<1-1/\alpha$ and negative if $t>1-1/\alpha$, and zero of $t=1-1/\alpha$. Hence the original function increases on $\left[0,1-1/\alpha \right]$ and decreases on $\left[1-1/\alpha ,\mathrm{\infty }\right)$, reaching its peak at $1-1/\alpha$.
All this is true if $\alpha >1$; if $\alpha <1$ then the function decreases on $\left[0,\mathrm{\infty }\right)$ and has its maximum at $t=0$.

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