Computing conditional log odds of dependent variable with only one observation at a particular independent variable

Question

I'm trying to understand logistic regression, and I keep getting hung up on the following point. Let $Y$ be the dependent variable taking its values in ${0,1}$ with a single independent variable $X$ in the model, and let $p(x) = P(Y=1|X=x)$. According to my understanding, what we're really doing is fitting a linear model to the log odds. That is, we seek the maximum likelihood estimation of $\beta_0$ and $\beta_1$ where $$\log(p(x)/(1-p(x))) = \beta_0 + \beta_1 x.$$ I'm confused, however, at how the left hand side is actually computed when this fit is made. In particular, I don't understand how $p(x)$ is computed in the case that there is only a single observation at a particular X value.

By way of example, in the iris data set included in R, suppose I wanted to fit a logistic regression to determine whether the species is setosa (1), or not (0) with the independent variable Sepal.Length. How does one compute $p(4.3)$ for the (singular) observation where Sepal.Length=4.3?

I came up against this when I tried to actually plot the linear function on the right side against the (observed) log odds computed from the data, and realized I had no idea how to compute the left side. Thanks for any help!

Answer 1

I think after you get you intercept and $\beta$ from the logistic regression you can directly calculate the probability for Sepal.Lenth=4.3

Here is the R code to do the logistic regression

iris$Species_new<-ifelse(iris$Species=='setosa',1,0) #recode the variable

myLogistic <- glm(Species_new ~ Sepal.Length, data = iris, family = "binomial")

results:

  #Call:  glm(formula = Species_new ~ Sepal.Length, family = "binomial",    data = iris)

  #Coefficients:
  #(Intercept)  Sepal.Length  
  #27.829        -5.176  

 #Degrees of Freedom: 149 Total (i.e. Null);  148 Residual
 #Null Deviance:      191 
 #Residual Deviance: 71.84        AIC: 75.84

Now your logistic regression model is:

$log\frac{p}{1-p}=27.829-5.176x \tag{1}$

$x$ is the Sepal.Length

Now Sepal.Length=$4.3$ plug this value into (1)

$log\frac{p}{1-p}=27.829-5.176*4.3$

It is easy to solve this equation:

$\frac{p}{1-p}=e^{(27.829-5.176*4.3)}$

Solve this we get $p=0.9962123$

Let us check what the computer get automatically,

p_i<-myLogistic$fit
p_i

You need to check the 14th data point for $p_i$ which is Sepal.Length=4.3

The $p$ is

0.9962154

It is almost the same as calculated by hand, the difference is only caused by different float point I think.