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

Question

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

Monastro3n · Accepted Answer

Step 1The 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 2Equating this to 0 you get   t  =  0  ,  (  α  −  1  )      /    λCheck the double derivative at these points to get the maximum.

Caylee Villegas · Accepted Answer

Step 1You&#039;re trying to maximize   t  ↦      t          α      −      1            e          −      α      t       on the interval   0  ≤  t  &amp;lt;  ∞. 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 2Since 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  &amp;lt;  1  −  1      /    α and negative if   t  &amp;gt;  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  &amp;gt;  1; if   α  &amp;lt;  1 then the function decreases on   [  0  ,  ∞  ) and has its maximum at   t  =  0.

If 1 leq alpha show show that the gamma density has a maximum at (alpha-1)/lambda.

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