Question

Really short question! Please help me to solve, thank you!

(30%)Q2 (Poisson regression): We collected n 15 independent count observations {Vi : 1,..., 15 and their corresponding covariates (i 1,..., 15). Assume the relationship between Vi and xỉ (for i-: 1, , 15) is yi ~ Poisson(A) and log(A) α+ßxi. Please 1) write down the likelihood function L(a, B|x, y) of the Poisson regression model; 2) derive the Newton method for maxmizing L(a, BIx, y); 3) implement the Newton method using R to get MLE of (α, β) i -0.30 0.32 0.41 0.62 -0.21 0.31 0.41 0.81 i 2 2 10 1 22 0 6 9 34 0.50 -0.21 -0.20 0.70 0.10 0.13 0.69 5 0 21 3 2 29

Answer 1

Here given that dependent (response) variable is Y, and we collected data n=15 with their corresponding independent variable X. Given that,

y_i ~ Poisson ($\lambda _{i}$) ...... i = 1, 2, ........... 15

And $Log(\lambda _{i})=\alpha + \beta x_{i}$

1) Let Likelihood function of the Poisson Regression model is as

$L(\alpha , \beta ;y_{i}) = \prod_{i=1}^{n} \frac{e^{-\lambda(x_{i})} [\lambda(x_{i})]^{y_{i}}}{y_{i}!}$

$= \prod_{i=1}^{n} \frac {e^{-e^{\alpha + \beta x_{i}}} [e^{\alpha+\beta x_{i}}]^{y_{i}}}{y_{i}!}$

The Log Likelihood function is as

$l(\alpha, \beta ; y_{i})= -\sum_{i=1}^{n}e^{\alpha + \beta x_{i}}+ \sum_{i=1}^{n} y_{i}(\alpha+\beta x_{i}) - \sum_{i=1}^{n}log(y_{i}!)$

2) The first order condition for a maximum is

$\frac{d}{d\lambda}l(\lambda; x_{1},......, x_{n}) = 0$

The first derivative of the log-likelihood with respect to the parameter $\lambda$ is

$\frac{d}{d\lambda}l(\lambda; x_{1},......, x_{n}) = \frac{d}{d\lambda} (-n\lambda - \sum_{j=1}^{n}ln(x_{j}!) + ln(\lambda)\sum_{j=1}^{n}x_{j})$

$=-n + \frac{1}{\lambda}\sum_{j=1}^{n}x_{j}$

Impose that the first derivative be equal to zero, and get

$\lambda =\frac{1}{n}\sum_{j=1}^{n}x_{j}$

3) By using R get MLE as

Here we get the value as

$\alpha = 0.7209$ and $\beta= 3.5419$

So the model is as

y = 0.7209 + 3.5419*x