I'm working with a zero inflated Poisson distribution that has the following pmf:


I would like to find the expectation of the following estimator:

$$\hat \lambda=\sum_i\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}$$

which is the sample mean for all non-zero observations. My result is as follows:

$$E[\hat \lambda]=\sum_iE\left[\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}\right]=\sum_i\sum_{y=0}^\infty\left[\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}\cdot f(y|w,\lambda)\right]$$

$$= \sum_i\sum_{y=0}^\infty\left[\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}\cdot wI[y=0]+\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}(1-w)\frac{e^{-\lambda}\lambda^{y}}{y!}\right]$$

The numerator of the first term in the square parentheses is always zero (opposing indicator functions); but here is where my troubles start, I'm not really sure how to handle the denominator on both terms, my approach was to continue as follows:

$$= \sum_i\sum_{y=1}^\infty\left[0+\frac{y_iI[y_i\neq 0]}{(\sum_iI[y_i\neq 0])}(1-w)\frac{e^{-\lambda}\lambda^{y}}{y!}\right]=\sum_i\sum_{y=1}^\infty\left[\frac{y_i}{n_{(y>0)}}(1-w)\frac{e^{-\lambda}\lambda^{y}}{y!}\right]=\frac{n(1-w)\lambda}{n_{(y>0)}}$$

Is this correct?

  • 1
    $\begingroup$ Not sure what the right expectation is, but the answer shouldn't contain the number of non-zero y's since that depends on the data. Also the answer should be greater than lambda (biased high) since all zeros are being ignored whether due to ZI or not. $\endgroup$
    – HStamper
    Jul 29, 2020 at 15:27
  • $\begingroup$ The estimator is not defined when all $y_i$'s are zero. $\endgroup$
    – Xi'an
    Jul 29, 2020 at 15:27
  • $\begingroup$ In that case you need to include a factor of $\mathcal{I}(y \ne 0)$ in the second term of $f.$ Note, too, that an estimator cannot estimate the "sample mean:" the sample mean is not a property of the distribution. It looks like you are trying to estimate the mean of the truncated distribution for $y\ne 0.$ Is that correct? $\endgroup$
    – whuber
    Jul 29, 2020 at 16:02
  • $\begingroup$ Why would we need an indicator function on the second term? The pmf seems proper without it (sums to 1, non-negative everywhere). Regarding the second part; I’m trying to estimate the parameter λ, which is the mean of the second component of a mixture distribution (the second component not being a truncated distribution, since $y=0$ can also arise from the second component) $\endgroup$
    – tvbc
    Jul 29, 2020 at 16:28
  • 1
    $\begingroup$ Without an indicator for the second term, the probability that $y=0$ is $w + (1-w)e^{-\lambda}:$ is this really what you intended? $\endgroup$
    – whuber
    Jul 29, 2020 at 20:01


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