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

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!

## 1 Answer

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$is 0.9962154  It is almost the same as calculated by hand, the difference is only caused by different float point I think. • Thanks for your response. But, isn't the quantity you've found above$\hat{p}$? My understanding may be incorrect here, so let me try to explain another way. In linear regression, the predicted$y$values are$\hat{y}$and the observed dependent variable values are$y_i$. Since logistic regression is in essence linear regression where the dependent variable is the log-odds (I know the coefficients are computed differently), how would one find the analog to the$y_i\$ values? Or, how would one compute the residuals if we view logistic regression as linear regression with logit response?
– Josh
Oct 3, 2015 at 18:30
• The principal with logistic regression is the same as linear regression. You can check David W. Hosmer's Applied Logistic Regression second Editon p147 goodness of fit. The Hosmer-Lemshow Test part. The observed value is the observed frequency with some characteristics and predicted value are predicted frequency with the same characteristics. Oct 4, 2015 at 4:51
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