# Estimate π of zero inflated Poisson

Can anyone show how to estimated the π of the zero-inflated-Poisson?

f(y)=π+(1−π)e^(−λ), if y=0;

=(1−π)λ^ye^(−λ)/y!, if y=1,2....

where π is the probability that the observation is zero by a binomial process and λ is the mean of the Poisson.

I use zeroinfl to get the estimated coefficients, but I do not know how to get π:

zeroinfl(formula = Lambda ~ Profil + Trace + Larg + Long + Reg + Vit + Racc + Momcorr | Profil + Trace + Larg + Long + Reg + Vit + Racc + Momcorr, data = LambdaData)

Count model coefficients (poisson with log link):

             Estimate     Std. Error   z value  Pr(>|z|)
(Intercept)  -5.320e+00    1.263e+00  -4.212   2.53e-05
Profil       5.560e-02    2.098e-01   0.265   0.790958
Trace        3.455e-01    1.013e-01   3.409   0.000652
Larg         3.630e-02    6.828e-02   0.532   0.594990
Long         5.674e-03    1.826e-02   0.311   0.756018
Reg         -2.828e-02    7.315e-01  -0.039   0.969163
Vit          4.495e-05    2.629e-03   0.017   0.986358
Racc         5.281e-01    4.964e-01   1.064   0.287449
Momcorr      3.699e-03    6.009e-04   6.156   7.48e-10


Zero-inflation model coefficients (binomial with logit link):

            Estimate   Std. Error   z value  Pr(>|z|)
(Intercept)  7.070353     1.990167   3.553   0.000381
Profil       0.678794     0.435449   1.559   0.119035
Trace        0.203185     0.250496   0.811   0.417291
Larg        -0.340889     0.191298  -1.782   0.074753
Long        -0.041204     0.065708  -0.627   0.530607
Reg         -3.507487     1.522579  -2.304   0.021242
Vit         -0.011700     0.005227  -2.238   0.025203
Racc        -0.336626     1.135956  -0.296   0.766973
Momcorr     -0.025258     0.012428  -2.032   0.042120


Number of iterations in BFGS optimization: 68 Log-likelihood: -2589 on 18 Df

• Welcome to Cross Validated. Can you clarify if you are trying to calculate it manually or obtain it from the package you are using? Note that if you are asking about how to obtain it from your statistical package that it is off-topic for this site, and may be closed – Marquis de Carabas Nov 9 '16 at 18:02
• I think there is a way to calculate it through the estimation results of "zeroinfl" function. – enzo liang Nov 10 '16 at 9:33

In zeroinfl() link functions are used to link the two parameters $\lambda$ and $\pi$ to the two sets of regressors $x$ and $z$. By default (as in your example), log and logit link functions are used, i.e., $\log(\lambda) = x^\top \beta$ and $\text{logit}(\pi) = z^\top \beta$. This is explained in vignette("countreg", package = "pscl") in Section 2.3. In your case $x = z$.
To obtain predictions for $\lambda$ and $\pi$ from a fitted zeroinfl object, obj say, you can use predict(obj, type = "count") and predict(obj, type = "zero"), respectively. Note that these are different from predict(obj, type = "response") (the expectation of the zero-inflated response) and predict(obj, type = "prob")[,1] (the probability of observing a zero).