Ashlyn Krause

Answered

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:

$g(t)=\frac{{\lambda}^{\alpha}{t}^{\alpha -1}{e}^{-\lambda t}}{{\int}_{0}^{\mathrm{\infty}}{t}^{\alpha -1}{e}^{-t}\phantom{\rule{thinmathspace}{0ex}}dt}$

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

So using this form of the gamma density function:

$g(t)=\frac{{\lambda}^{\alpha}{t}^{\alpha -1}{e}^{-\lambda t}}{{\int}_{0}^{\mathrm{\infty}}{t}^{\alpha -1}{e}^{-t}\phantom{\rule{thinmathspace}{0ex}}dt}$

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

Answer & Explanation

Monastro3n

Expert

2022-07-17Added 15 answers

Step 1

The denominator is a constant as you are integrating it over t. So it is same as maximizing only the numerator. And $\frac{d({t}^{\alpha -1}{e}^{-\alpha t})}{dx}=(\alpha -1){t}^{\alpha -2}{e}^{-\lambda t}-\lambda {t}^{\alpha -1}{e}^{-\lambda t}$.

Step 2

Equating this to 0 you get $t=0,(\alpha -1)/\lambda $

Check the double derivative at these points to get the maximum.

The denominator is a constant as you are integrating it over t. So it is same as maximizing only the numerator. And $\frac{d({t}^{\alpha -1}{e}^{-\alpha t})}{dx}=(\alpha -1){t}^{\alpha -2}{e}^{-\lambda t}-\lambda {t}^{\alpha -1}{e}^{-\lambda t}$.

Step 2

Equating this to 0 you get $t=0,(\alpha -1)/\lambda $

Check the double derivative at these points to get the maximum.

Caylee Villegas

Expert

2022-07-18Added 2 answers

Step 1

You're trying to maximize $t\mapsto {t}^{\alpha -1}{e}^{-\alpha t}$ 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:

$\mathrm{log}\left({t}^{\alpha -1}{e}^{-\alpha t}\right)=(\alpha -1)\mathrm{log}t-\alpha t.$

The derivative of this with respect to t is $\frac{\alpha -1}{t}-\alpha =\alpha \frac{1-(1/\alpha )-t}{t}.$.

Step 2

Since the denominator is positive, the question now is only when the numerator is positive, negative, and zero. The numerator $1-(1/\alpha )-t$ 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 $[0,1-1/\alpha ]$ and decreases on $[1-1/\alpha ,\mathrm{\infty})$, reaching its peak at $1-1/\alpha $.

All this is true if $\alpha >1$; if $\alpha <1$ then the function decreases on $[0,\mathrm{\infty})$ and has its maximum at $t=0$.

You're trying to maximize $t\mapsto {t}^{\alpha -1}{e}^{-\alpha t}$ 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:

$\mathrm{log}\left({t}^{\alpha -1}{e}^{-\alpha t}\right)=(\alpha -1)\mathrm{log}t-\alpha t.$

The derivative of this with respect to t is $\frac{\alpha -1}{t}-\alpha =\alpha \frac{1-(1/\alpha )-t}{t}.$.

Step 2

Since the denominator is positive, the question now is only when the numerator is positive, negative, and zero. The numerator $1-(1/\alpha )-t$ 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 $[0,1-1/\alpha ]$ and decreases on $[1-1/\alpha ,\mathrm{\infty})$, reaching its peak at $1-1/\alpha $.

All this is true if $\alpha >1$; if $\alpha <1$ then the function decreases on $[0,\mathrm{\infty})$ and has its maximum at $t=0$.

Most Popular Questions