Consider random variable Y with a Poisson distribution: P ( y | θ ) =...

spockmonkey40

spockmonkey40

Answered

2022-07-07

Consider random variable Y with a Poisson distribution:
P ( y | θ ) = θ y e θ y ! , y = 0 , 1 , 2 , , θ > 0
P ( y | θ ) = θ y e θ y ! , y = 0 , 1 , 2 , , θ > 0
Mean and variance of Y given θ are both equal to θ. Assume that i = 1 n y i > 1.
If we impose the prior p 1 θ , then what is the Bayesian posterior mode?
I was able to calculate the likelihood and the posterior, but I'm having trouble calculating the mode so I'm wondering if I got the right posterior:
P ( θ | y ) = l i k e l i h o o d p r i o r
P ( θ | y ) ( θ i = 1 n y i e n θ ) ( θ 1 )
P ( θ | y ) θ ( i = 1 n y i ) 1 e n θ

Answer & Explanation

Marisol Morton

Marisol Morton

Expert

2022-07-08Added 13 answers

The posterior mode is just the maximizing value of the posterior, so this is essentially just a calculus problem. You have already correctly derived the posterior kernel:
π ( θ | y ) exp ( ( n y ¯ 1 ) ln θ n θ ) .
So the log-posterior can be written as:
F y ( θ ) ln π ( θ | y ) = ( n y ¯ 1 ) ln θ n θ + const .
We can maximise this via ordinary calculus techniques. Differentiating with respect to θ gives:
d F y d θ ( θ ) = n y ¯ 1 θ n d 2 F y d θ 2 ( θ ) = n y ¯ 1 θ 2 .
You are told that n y ¯ > 1 so the second derivative of the objective function is negative. This means that the objective function is strictly concave, so the maximizing value occurs at the unique critical point:
0 = d F y d θ ( θ ^ ) = n y ¯ 1 θ ^ n θ ^ = y ¯ 1 n .
So in this case we have mode mode  π ( θ | y ) = y ¯ 1 / n (which is strictly positive since y ¯ > 1 / n).

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