# Using the equation of a generalized linear mixed effects model to predict x values

Very similar to this question Find the equation from generalized linear model output, I want to derive an equation for the logistic regression that includes the estimates of a random effect.

I've included the following code for reference:

Trial=c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3)
Time=c(2.0, 6.0, 9.0, 12.0, 15.0, 18.0, 21.0, 24.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 1.5, 3.0, 4.5, 6.0, 39.0)
Alive=c(10, 0,  0,  0,  0,  0,  0,  0,  6,  2,  8,  1,  0,  0,  4,  6,  1,  2, 0)
Dead=c(0, 10,  6, 10, 10, 10,  7, 10,  0,  8,  1,  9, 10, 10,  5,  0,  8,  6, 10)
ostrinaA.glmm<- glmer(cbind(Alive, Dead)~Time+(1|Trial), family = binomial(link="logit"))


Retrieving the coefficients;

> ranef(ostrinaA.glmm)
$Trial (Intercept) 1 0.13701996 2 -0.17639754 3 0.04042084 > coef(ostrinaA.glmm)$Trial
(Intercept)       Time
1    2.778796 -0.9009119
2    2.465379 -0.9009119
3    2.682197 -0.9009119

attr(,"class")
[1] "coef.mer"
> fixef(ostrinaA.glmm)
(Intercept)        Time
2.6417764  -0.9009119
> summary(ostrinaA.glmm)
Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod']
Family: binomial  ( logit )
Formula: cbind(Alive, Dead) ~ Time + (1 | Trial)
Data: ostrinaA

AIC      BIC   logLik deviance df.resid
69.7     72.5    -31.8     63.7       16

Scaled residuals:
Min       1Q   Median       3Q      Max
-3.07128 -0.78873 -0.01476  0.56018  2.70666

Random effects:
Groups Name        Variance Std.Dev.
Trial  (Intercept) 0.06756  0.2599
Number of obs: 19, groups:  Trial, 3

Fixed effects:
Estimate Std. Error z value    Pr(>|z|)
(Intercept)   2.6418     0.6498   4.065 0.000047964 ***
Time         -0.9009     0.1695  -5.316 0.000000106 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
(Intr)
Time -0.858


If this were simply a generalized linear model with the logit link function I believe my equation would read;

However, this does not include the random effect. How would I include the effect of trial in my equation? Ultimately, I'd like to use this equation to predict an x value (Time) where the response (Alive/Dead) is equal to 50% (LT50).

## 1 Answer

The random effect is not a number, but a random variable, with mean 0, and standard deviation $\sigma=0.2599$ in this case. It follows a normal distribution. Something you can do is simulate a large sample of the random effect, since you know what distribution it follows, and then average the predictions, like a Monte Carlo procedure. But, also, it is likely that the R library you are using already has a prediction function for this type of regression.

If you are asking about the formal way of writing the equation, it would be $$P(y=1|x,u)=\frac{1}{1+e^{-(2.6418-0.9009x+u)}},$$ where $u\sim N(0,0.2599^2)$.

• I am more concerned about making a prediction with the equation than the formal way of writing but your answer was still very helpful, thank you! The prediction I'm trying to make is not of the response but rather the treatment (Time) at a response of 50% mortality (Alive=5, Dead=5)..Makes it tricky. – hamilthj Mar 14 '17 at 5:43
• If I were to have trial as a fixed effect rather than a random effect, would the equation be similar but with u as a 2nd parameter instead of random effect? – hamilthj Apr 4 '17 at 17:30
• @hamilthj It would, yes. – Anna SdTC Apr 5 '17 at 0:36
• In the model output, what does the intercept represent. In this regression, the y intercept is near 1 and there is no x intercept. – hamilthj May 4 '17 at 20:25
• @hamilthj I don't understand your question. – Anna SdTC May 10 '17 at 2:49