0
$\begingroup$

any help with these questions would be much appreciated, thank you.

Assume that the response variable $Y$ in a regression problem is a Bernoulli random variable, that is $Y$ ~ $Be(\pi(\beta'x))$, where $\pi(\beta'x)$ is a logistic function, $\beta'x=\beta_0+\beta_1x_1+\beta_2x_2+...+\beta_px_p$ and $x=(1,x_1,x_2,...,x_p)$, i.e., $Y$ follows a logistic regression.
Let $(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$ be a data set of independent samples, where $y_i$ ∈ {$0,1$} and $x_i=(1,x_{i1},x_{i2},...,x_{ip}), i=1,2,...,n$.

a) Show that for all real $\beta$, the log likelihood function $l(\beta)$ can be written as
$l(\beta)=∑y_i\beta'x_i-∑ln(1+exp(\beta'x))$.

b) Find the partial derivatives $∂/∂β_0$ $l(\beta)$ and $∂^2/∂β^2_0$ $l(\beta)$ in the form they would appear in a recursive algorithm like Newton-Raphson for finding the maximum likelihood estimate.

$\endgroup$

1 Answer 1

0
$\begingroup$
  1. $y_i \sim \mathcal{B}er(p_i)$, and $p_i \equiv p(x_i'\beta_i) = (1 + \exp\{-x_i'\beta\})^{-1} $ and $$ \ln p(x_i'\beta) =\ln \left(\frac{e^{x_i'\beta}}{1 +e^{x_i'\beta}} \right) = x_i'\beta - \ln (1+e^{x_i'\beta}) $$ thus $$ \mathcal{L}(\beta)=\prod_{i=1}^n p(x_i'\beta)^{y_i}(1-p(x_i'\beta))^{1-y_i} $$ $$ l(\beta) = \sum y_i \ln p(x_i'\beta_i) + \sum (1- y_i)\ln( 1- p(x_i'\beta_i)) $$ $$ l(\beta) = \sum y_i x_i'\beta - \sum y_i \ln(1 + e^{x_i'\beta}) $$

  2. Show that $$ p(x_i'\beta)'_{\beta_0} = p_i(1-p_i) = \frac{1}{1+e^{x_i'\beta}}\times\frac{e^{x_i'\beta}}{1+e^{x_i'\beta}} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .