If 1 ≤ α show show that the gamma density has a maximum at α...

Ashlyn Krause

Ashlyn Krause

Answered

2022-07-16

If 1 α show show that the gamma density has a maximum at α 1 λ .
So using this form of the gamma density function:
g ( t ) = λ α t α 1 e λ t 0 t α 1 e t d t
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

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 d ( t α 1 e α t ) d x = ( α 1 ) t α 2 e λ t λ t α 1 e λ t .
Step 2
Equating this to 0 you get t = 0 , ( α 1 ) / λ
Check the double derivative at these points to get the maximum.
Caylee Villegas

Caylee Villegas

Expert

2022-07-18Added 2 answers

Step 1
You're trying to maximize t t α 1 e α t on the interval 0 t < . 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:
log ( t α 1 e α t ) = ( α 1 ) log t α t .
The derivative of this with respect to t is α 1 t α = α 1 ( 1 / α ) 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 / α ) t is positive if t < 1 1 / α and negative if t > 1 1 / α, and zero of t = 1 1 / α. Hence the original function increases on [ 0 , 1 1 / α ] and decreases on [ 1 1 / α , ), reaching its peak at 1 1 / α.
All this is true if α > 1; if α < 1 then the function decreases on [ 0 , ) and has its maximum at t = 0.

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