How to create a logistic regression model based on a two way table

Question

question

Given a table such as : $$ \begin{array}{ll|ll} & A & {} \\ & & 0 & 1 \\ \hline B& 1 & 44 & 27 \\ &0 & 443 & 95 \end{array} $$

If I wanted to develop a logistic regression model that included $A$ only, how would I go about that.

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following Maarten Buis example

Following this but predicting $A$ instead of $B$

$$ \ln( \text{Odds}(A=1 \vert B)) = \beta_0 + \beta_1 B $$

Then $\beta_0$ is found from the log odds of $A$ being 1 when $B=0$, which is $\ln( 95/443) = -1.54$. So the odds of $A$ given $B=0$ are $95/443$.

Then for $\beta_1$ we have the log odds of $A$ when $B=1$ which is $\ln( 27 / 44 ) = -0.49$. Meaning that for a unit increase in $B$ the odds change by $(27/44)\div(95/443) \approx 2.9$. About three times larger if we predict $A$ from $B$, which matches the change in predicting $B$ from $A$.

Answer 1

You need to choose whether $A$ or $B$ is your dependent/explained/left-hand-side/$y$-variable. Let's say $B$ is the variable you want to explain, and $A$ is the variable with which you explain $B$. So now you can write the regression as:

$\ln(Odds(B=1|A))= \beta_0+\beta_1A$

So $\beta_0$ is a constant, so it is the log odds of $B$ being 1 when all $x$-variables (in this case just $A$) equal 0. The odds is the number of successes per failure, which in your example is $\frac{44}{443}\approx0.10$. Taking the log of that will give us $\beta_0$, which for your table is approximately $-2.3$.

$\beta_1$ is the log odds ratio. The odds ratio is, as the name suggest, a ratio of odds. It answers the question, by what factor does the odds change for a unit increase in $x$. The odds was $\frac{44}{443}$ when $A$ was 0, and the odds is $\frac{27}{95}$ when $A$ is 1, so odds changes by a factor $\frac{\frac{27}{95}}{\frac{44}{443}} = \frac{27\times443}{95\times44}\approx2.9$. So the odds of $B=1$ is almost three times larger for the group $A=1$ compared to the group $A=0$. Taking the log of the odds ratio will give is $\beta_1$, in this case approximately $1.1$

These manual computation were possible because this is a so called saturated model; the model will exactly reproduce the cell counts. If, for example, we added a second explanatory variable but not the interaction, then this would no longer be a saturated model, and we would have to use an iterative algorithm to find the maximum likelihood estimates of the parameters.

In your question you referred to just using $A$, and shown the marginal distribution of $A$. That would be equivalent to a logistic regression with only a constant and no explanatory variables. This too is a saturated model, and the constant in that model is the log of the odds you estimated.

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Edit, response to edit in original question

What you found is correct: the odds ratio is symmetric. This property is for example used in case control studies.

Answer 2

@Maarten Buis has done an excellent job explaining your logistic regression questions, but I figured I'd supplement his answer with some sample R code so you can see the calculations performed in a statistical analysis program.

In the first part of your question, you asked about the computation of:

$$ log( odds(B=1|A) = \beta_0 + \beta_1 A $$

We can get this model ( log-odds of $B=1$ given $A$) as follows:

> A<-c(rep(0,44+443), rep(1, 27+95))
> B<-c(rep(1, 44), rep(0, 443), rep(1, 27), rep(0, 95))
> 
> #view dummy table in table form
> table(B,A)
   A
B     0   1
  0 443  95
  1  44  27
>      
> #run log(odds(B=1|A))
> mylogit_BgivenA <- glm(B ~ A, family = "binomial")
> mylogit_BgivenA

Call:  glm(formula = B ~ A, family = "binomial")

Coefficients:
(Intercept)            A  
     -2.309        1.051  

You can see from the coefficients the -2.309 here, corresponds to the log odds of B being equal to 1 given, all the values of A are equal to 0. This also corresponds to Maarten's approximate value of -2.3.

Moving on to your second question, we can find the probabilities and log odds of A by regressing A onto a constant value of 1:

> mylogit_Aonly <- glm(A ~1, family = "binomial")
> 
> mylogit_Aonly

Call:  glm(formula = A ~ 1, family = "binomial")

Coefficients:
(Intercept)  
     -1.384  

> 
> exp_b0<-exp(-1.384)
> #probability of event A given no explanatory variables
> #calculate the probability of p_A
> exp_b0/(1+exp_b0)

[1] 0.2003674

Lastly, note that $log(odds(B=1|A)) \ne log(odds(A=1|B))$. You can verify this by running another logistic regression and comparing the results with the first output above:

> #run log(odds(A=1|B))
> mylogit_AgivenB <- glm(A ~ B, family = "binomial")
> mylogit_AgivenB

Call:  glm(formula = A ~ B, family = "binomial")

Coefficients:
(Intercept)            B  
     -1.540        1.051  

Note that the value of the intercept (-1.54) is different than that calculated in the first output shown (-2.309).