Poisson MLE and Limiting Distribution

[mlarson9000]

Homework Statement

Let yi denote the number of times individual i buys tobacco in a given month.
Suppose a random sample of N individuals is available, for which we observe values
0,1,2,... for yi.
Let xi be an observed characterisitc of these individuals (for example, gender). If we assume that for a given xi, yi has a Poisson distribution with mean
λi = exp ( β1 + β 2xi).
That is, the distribution function is:

Pr (yi = y |xi) =[exp(-λi)λi^y]/y! y = 0,1,2,...

(a) Obtain B_mle and derive its limiting distribution.
(b) Now suppose that yi does not take Poisson distribution. However, the orthogonality
condition holds:
E (yi - exp (β 1 + β 2xi) | xi) = 0:
Propose a consistent estimator for and derive its limiting distribution.

Homework Equations

The Attempt at a Solution

I'm not sure how to approach this. Do I just plug exp(β1+β2xi) into λi, and derive the MLE, or am I supposed to do something else? When I tried it that way, I get

dL/dβ1=ny-Ʃexp(β1+β2xi)=0

dL/dβ2=yƩxi-Ʃxi[exp(β1+β2xi)]=0

And I'm not sure how to solve for β1 and β2 with this.

 

[mighty2000]

Thank you for your post. To solve this problem, we can use the method of maximum likelihood estimation (MLE). This method is commonly used in statistics to estimate the parameters of a probability distribution based on a given set of data. In this case, we are trying to estimate the parameters β1 and β2 based on a given sample of individuals and their characteristics.

To obtain the MLE, we need to maximize the likelihood function L(β1, β2) with respect to the parameters β1 and β2. This function is defined as the product of the individual probability densities for each observation in the sample. In this case, the individual probability density is given by:

Pr(yi = y | xi) = [exp(λi - λi)λiy]/y!

Substituting the given expression for λi, we get:

Pr(yi = y | xi) = [exp(β1 + β2xi - exp(β1 + β2xi))exp(β1 + β2xi)y]/y!

Taking the logarithm of both sides, we get:

ln L(β1, β2) = Σ[β1 + β2xi - exp(β1 + β2xi) + yln(exp(β1 + β2xi)) - ln(y!)].

To maximize this function, we need to take the partial derivatives with respect to β1 and β2 and set them equal to 0. This will give us the maximum likelihood estimates for β1 and β2. Taking the partial derivative with respect to β1, we get:

∂ln L(β1, β2)/∂β1 = Σ[1 - exp(β1 + β2xi) + yexp(β1 + β2xi)] = 0.

Solving for β1, we get:

β1 = ln(Σ[y - exp(β1 + β2xi)]/N).

Similarly, taking the partial derivative with respect to β2, we get:

∂ln L(β1, β2)/∂β2 = Σ[xi - exp(β1 + β2xi) + yxiexp(β1 + β2xi)] = 0.

Solving for β2, we get:

β2 = Σ[(y - exp(β1 + β2xi))xi]/Σ[xi^