How to get probability from GLM output

Question

I\'m extremely stuck at the moment as I am trying to figure out how to calculate the probability from my glm output in R. I know the data is very insignificant

Answer 1

The dependent variable in a logistic regression is a log odds ratio. We'll illustrate how to interpret the coefficients with the space shuttle autolander data from the MASS package.

After loading the data, we'll create a binary dependent variable where:

1 = autolander used, 
0 = autolander not used. 

We will also create a binary independent variable for shuttle stability:

1 = stable positioning
0 = unstable positioning. 

Then, we'll run glm() with family=binomial(link="logit"). Since the coefficients are log odds ratios, we'll exponentiate them to turn them back into odds ratios.

library(MASS)
str(shuttle)
shuttle$stable <- 0
shuttle[shuttle$stability =="stab","stable"] <- 1
shuttle$auto <- 0
shuttle[shuttle$use =="auto","auto"] <- 1

fit <- glm(use ~ factor(stable),family=binomial(link = "logit"),data=shuttle) # specifies base as unstable

summary(fit)
exp(fit$coefficients)

...and the output:

> fit <- glm(use ~ factor(stable),family=binomial(link = "logit"),data=shuttle) # specifies base as unstable
> 
> summary(fit)

Call:
glm(formula = use ~ factor(stable), family = binomial(link = "logit"), 
data = shuttle)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-1.1774  -1.0118  -0.9566   1.1774   1.4155  

Coefficients:
                  Estimate Std. Error z value Pr(>|z|)  
(Intercept)      4.747e-15  1.768e-01   0.000   1.0000  
factor(stable)1 -5.443e-01  2.547e-01  -2.137   0.0326 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 350.36  on 255  degrees of freedom
Residual deviance: 345.75  on 254  degrees of freedom
AIC: 349.75

Number of Fisher Scoring iterations: 4

> exp(fit$coefficients)
    (Intercept) factor(stable)1 
      1.0000000       0.5802469 
> 

The intercept of 0 is the log odds for unstable, and the coefficient of -.5443 is the log odds for stable. After exponentiating the coefficients, we observe that the odds of autolander use under the condition of an unstable shuttle 1.0, and are multiplied by .58 if the shuttle is stable. This means that the autolander is less likely to be used if the shuttle has stable positioning.

Calculating probability of autolander use

We can do this in two ways. First, the manual approach: exponentiate the coefficients and convert the odds to probabilities using the following equation.

p = odds / (1 + odds) 

With the shuttle autolander data it works as follows.

# convert intercept to probability
odds_i <- exp(fit$coefficients[1])
odds_i / (1 + odds_i)
# convert stable="stable" to probability
odds_p <- exp(fit$coefficients[1]) * exp(fit$coefficients[2])
odds_p / (1 + odds_p)

...and the output:

> # convert intercept to probability
> odds_i <- exp(fit$coefficients[1])
> odds_i / (1 + odds_i)
(Intercept) 
        0.5 
> # convert stable="stable" to probability
> odds_p <- exp(fit$coefficients[1]) * exp(fit$coefficients[2])
> odds_p / (1 + odds_p)
(Intercept) 
  0.3671875 
>

The probability of autolander use when a shuttle is unstable is 0.5, and decreases to 0.37 when the shuttle is stable.

The second approach to generate probabilities is to use the predict() function.

# convert to probabilities with the predict() function
predict(fit,data.frame(stable="0"),type="response")
predict(fit,data.frame(stable="1"),type="response")

Note that the output matches the manually calculated probabilities.

> # convert to probabilities with the predict() function
> predict(fit,data.frame(stable="0"),type="response")
  1 
0.5 
> predict(fit,data.frame(stable="1"),type="response")
        1 
0.3671875 
> 

Applying this to the OP data

We can apply these steps to the glm() output from the OP as follows.

coefficients <- c(-1.1455,-0.1322)
exp(coefficients)
odds_i <- exp(coefficients[1])
odds_i / (1 + odds_i)
# convert nonRSEvents = 1 to probability
odds_p <- exp(coefficients[1]) * exp(coefficients[2])
odds_p / (1 + odds_p)
# simulate up to 10 nonRSEvents prior to RS
coef_df <- data.frame(nonRSEvents=0:10,
                  intercept=rep(-1.1455,11),
                  nonRSEventSlope=rep(-0.1322,11))
coef_df$nonRSEventValue <- coef_df$nonRSEventSlope * 
coef_df$nonRSEvents
coef_df$intercept_exp <- exp(coef_df$intercept)
coef_df$slope_exp <- exp(coef_df$nonRSEventValue)
coef_df$odds <- coef_df$intercept_exp * coef_df$slope_exp
coef_df$probability <- coef_df$odds / (1 + coef_df$odds)
# print the odds & probabilities by number of nonRSEvents
coef_df[,c(1,7:8)]

...and the final output.

> coef_df[,c(1,7:8)]
   nonRSEvents    odds probability
1            0 0.31806     0.24131
2            1 0.27868     0.21794
3            2 0.24417     0.19625
4            3 0.21393     0.17623
5            4 0.18744     0.15785
6            5 0.16423     0.14106
7            6 0.14389     0.12579
8            7 0.12607     0.11196
9            8 0.11046     0.09947
10           9 0.09678     0.08824
11          10 0.08480     0.07817
> 

Answer 2

If you want to predict the probability of response for a specified set of values of the predictor variable:

pframe <- data.frame(NonRSevents_before1stRS=4)
predict(fitted_model, newdata=pframe, type="response")

where fitted_model is the result of your glm() fit, which you stored in a variable. You may not be familiar with the R approach to statistical analysis, which is to store the fitted model as an object/in a variable, then apply different methods to it (summary(), plot(), predict(), residuals(), ...)

• This is obviously only a made-up example: I don't know if 4 is a reasonable value for the NonRSevents_before1stRS variable)
• you can specify more different values to do predictions for at the same time (data.frame(NonRSevents_before1stRS=c(4,5,6,7,8)))
• if you have multiple predictors, you have to specify some value for every predictor for every prediction, e.g. data.frame(x=4:8,y=mean(orig_data$y), ...)

If you want the predicted probabilities for the observations in your original data set, just predict(fitted_model, type="response")

You're correct that inv.logit() (from a bunch of different packages, don't know which you're using) or plogis() (from base R, essentially the same) will translate from the logit or log-odds scale to the probability scale, so

plogis(predict(fitted_model))

would also work (predict provides predictions on the link-function [in this case logit/log-odds] scale by default).