7
$\begingroup$

I know that the density looks like this: $$P(Y=y) = \frac{e^{-\lambda} \lambda^y}{y!(1-e^{-\lambda})}$$ and from Wikipedia that the mean and variance are

$$\operatorname E(Y) = \frac{\lambda}{1-e^{-\lambda}}$$

$$ \operatorname{Var}(Y) = \frac{\lambda + \lambda^2}{1-e^{- \lambda}} - \frac{\lambda^2}{(1-e^{- \lambda})^2}.$$

But how have the mean and variance been derived?

$\endgroup$
2
  • $\begingroup$ Do you also seek for usual Poisson mean/variance derivation? $\endgroup$
    – gunes
    Apr 27, 2019 at 17:42
  • 1
    $\begingroup$ If you only get a report when something is counted, then the reports are zero-truncated $\endgroup$
    – Henry
    Apr 27, 2019 at 22:04

3 Answers 3

4
$\begingroup$

$\newcommand{\e}{\operatorname E}\newcommand{\v}{\operatorname{var}}$This appears to include one or more applications of the zero-truncated Poisson distribution in applied statistics.

Suppose $X\sim\operatorname{Poisson}(\lambda)$ and $Y= \begin{cases} 1 & \text{if } X=0, \\ 0 & \text{otherwise.} \end{cases}$

The zero-truncated Poisson distribution is the conditional distribution of $X$ given then event $Y=0.$ So we seek $\e(X\mid Y=0)$ and $\v(X\mid Y=0).$

First, apply the law of total expectation: \begin{align} & \lambda =\e(X) = \e(\e(X\mid Y)) \\[6pt] = {} & \e(X\mid Y=0)\Pr(Y=0) + \e(X\mid Y=1)\Pr(Y=1) \\[6pt] = {} & \e(X\mid Y=0)(1-e^{-\lambda}) + 0. \end{align} So we have $$ \lambda =\e(X\mid Y=0)(1-e^{-\lambda} ) $$ and therefore $$ \e(X\mid Y=0) = \frac \lambda {1-e^{-\lambda}}, $$ or, if you like, $$ \e(X\mid Y=0) = \frac{\lambda e^\lambda}{e^\lambda -1}. $$ Next apply the law of total variance: $$ \lambda = \v(X) = \v(\e(X\mid Y)) + \e(\v(X\mid Y)). \tag 0 $$ And \begin{align} \e(\v(X\mid Y)) = {} & \v(X\mid Y=0)\Pr(Y=0) \\ & {} + \v(X\mid Y=1)\Pr(Y=1) \\[6pt] = {} & \v(X\mid Y=0) (1-e^{-\lambda}) + 0. \tag 1 \end{align} \begin{align} \v(\e(X\mid Y)) = {} & \v\left.\begin{cases} 0 & \text{if } Y = 1, \\ \lambda/(1-e^{-\lambda}) & \text{if } Y=0 \end{cases} \right\} \\[10pt] = {} & \Pr(Y=0)\cdot\Pr(Y=1) \cdot \left( \frac \lambda {1-e^{-\lambda}} \right)^2 \\[8pt] = {} & (1-e^{-\lambda}) e^{-\lambda} \cdot \left( \frac \lambda {1-e^{-\lambda}} \right)^2 \tag 2 \end{align} So the sum of the two quantities on lines $(1)$ and $(2)$ above is equal, as stated on line $(0),$ to $\lambda.$ From that you can deduce $\v(X\mid Y=0),$ the variance of the truncated Poisson distribution.

$\endgroup$
3
  • $\begingroup$ How do you know $Pr(Y=0)=(1-e^{-\lambda})$ without knowing the distribution of $Y$? $\endgroup$
    – Deep North
    Apr 27, 2019 at 22:00
  • 2
    $\begingroup$ @DeepNorth : I said $$X\sim\operatorname{Poisson}(\lambda)$$ and $$Y= \begin{cases} 1 & \text{if } X=0, \\ 0 & \text{otherwise.} \end{cases}$$ That tells you the distribution of $Y. \qquad$ $\endgroup$ Apr 28, 2019 at 2:02
  • 1
    $\begingroup$ @DeepNorth : $\ldots\ldots\,$in particular, it tells you that $\Pr(Y=1) = \Pr(X=0)$ and $\Pr(Y=0) = 1 - \Pr(X=0). \qquad$ $\endgroup$ Apr 28, 2019 at 2:36
3
$\begingroup$

Note that your PMF is $p(y)=\frac{f(y)}{1-e^{-\lambda}}$, for $y>0$, where $f(y)$ is the PMF of normal Poisson. Then,

$$\operatorname E[Y]=\sum_{y=1}^\infty y \frac{f(y)}{1-e^{-\lambda}}=\frac{1}{1-e^{-\lambda}}\sum_{y=1}^\infty y f(y)=\frac{1}{1-e^{-\lambda}}\underbrace{\sum_{y=0}^\infty yf(y)}_{\lambda\,\,\text{(i.e. Poisson mean)}}=\frac{\lambda}{1-e^{-\lambda}}$$

Similarly, we can find $E[Y^2]$ as $\frac{\lambda+\lambda^2}{1-e^{-\lambda}}$, yielding a variance equation (yours have an extra $\pm$ in the first summand):

$$\operatorname{var}(Y)=\frac{\lambda+\lambda^2}{1-e^{-\lambda}}-\frac{\lambda^2}{(1-e^{-\lambda})^2}$$

$\endgroup$
2
$\begingroup$

Concerning the use, one common situation that would make many people consider some kind of zero-truncated model would be any count data where zero counts are impossible (because, for whatever reason, they cannot be observed).


Concerning the derivation of the moments, you need to know three basic facts about Poisson$(\lambda)$ distributions--which I will highlight as we go--but only the most trivial mathematical calculation is needed, which appears at the very end.

(It's always better to find a clear, simple derivation because (i) it will provide insight and (ii) it reduces the chance of error. This is one basis for applying the Principle of Mathematical Laziness.)

Notice, then, that the formulas for moments of mixture distributions at https://stats.stackexchange.com/a/16609/919 apply directly--without any modification, despite having a negative weight!--because the zero-truncated Poisson distribution is a mixture of an atom at zero and a Poisson distribution with weights $p_0$ and $p_1,$ respectively, determined by two simple considerations:

  1. They must sum to unity, $p_0+p_1 = 1.$

  2. $p_0$ must be the amount of probability we need to add to cancel all chance that $X=0$ (where $X$ has a Poisson$(\lambda)$ distribution). Because $\Pr(X=0) = e^{-\lambda}$ (Fact 1 about Poisson distributions), the remaining probability is $1-e^{-\lambda},$ requiring a subsequent renormalization of the probability distribution to sum to unity: that is, everything must be divided by the remaining probability. Consequently

$$p_0 = -\frac{e^{-\lambda}}{1-e^{-\lambda}};\quad p_1 = \frac{1}{1 - e^{-\lambda}}.$$

Notice that $p_0$ must be negative. That doesn't change any of the formulas I will quote for moments of mixtures, as you can check if you wish.

Because the mean and variance of the Poisson$(\lambda)$ distribution are both $\lambda$ (Facts 2 and 3 about Poisson distributions), implying the second moment of the Poisson distribution is $\lambda+\lambda^2$), you can immediately write, without further thought or calculation,

$$E[Y] = p_0(0) + p_1(\lambda)$$

and

$$E[Y^2] = p_0(0^2) + p_1(\lambda + \lambda^2),$$

giving (still with no calculation whatsoever)

$$E[Y] = p_1\lambda$$

and

$$\operatorname{Var}(Y) = E[Y^2] - E[Y]^2 = p_1(\lambda + \lambda^2) - (p_1\lambda)^2.\tag{*}$$


Let's review. Suppose $X$ is any random variable. Write $X_*$ for its zero truncation (that is, $0$ is eliminated as a possible value.) Let $\mu_k$ refer to the $k^\text{th}$ moment, so (for instance) $\mu_k(X) = E[X^k].$ Looking back over this analysis, we have obtained (a) a general formula for moments of zero-truncated distributions,

$$\mu_k(X_*) = \frac{1}{1 - \Pr(X=0)}\mu_k(X),$$

and (b) a specific formula for zero-truncated Poisson distributions,

$$\mu_k(X_*) = \frac{1}{1 - e^{-\lambda}}\mu_k(X).$$

From the latter, $(*)$ becomes

$$\operatorname{Var}(X_*) = \frac{\lambda + \lambda^2}{1-e^{-\lambda}} - \left(\frac{\lambda}{1-e^{-\lambda}}\right)^2.$$

Only at this point should you feel any need to do some algebra. Indeed, the square of the right hand fraction is the square of its numerator divided by the square of its denominator, yielding precisely the formula quoted in the question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.