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For each student, the probability of both falling asleep and dreaming about mathematics is $(1-p)q$. These indicators are independent across students, so the same Poisson thinning calculation gives, for the requested count $D$,Explicitly $G_D(s)=\exp (\lambda (1-p)q(s-1))$, the probability generating function of that Poisson distribution.

Let $I_i$ indicate that student $i$ stays awake. These indicators have independent Bernoulli distributions with parameter $p$, and $W={\sum }_{i=1}^M{I}_i$. The Poisson distribution has probability generating functionThe probability generating function of an independent random sum now givesRecognizing the Poisson distribution,This is Poisson thinning.

Given $M=m$, there are exactly $m$ independent opportunities to stay awake, each with probability $p$. HenceThe standard binomial distribution formulas are $\mathbb{E}[W\mid M]=pM$ and $\mathrm{var}(W\mid M)=Mp(1-p)$. The Poisson distribution formulas are $\mathbb{E}[M]=\lambda$, $\mathrm{var}(M)=\lambda$, and therefore $\mathbb{E}[M^2]={\lambda }^2+\lambda$. By the law of total expectation,Using the definition of covariance,Thus $W$ and $M$ are not independent, even though the separate awake and asleep counts produced by Poisson thinning are independent.

By definition, the probability generating function isSubstituting the Poisson distribution and evaluating the exponential series givesThe series converges for every finite complex $s$; in particular it gives the probability generating function on $0\le s\le 1$.

The exponential series gives ${\sum }_{n=0}^{\infty }{\lambda }^n/n!=e^{\lambda }$. Thus the normalization constant for the Poisson distribution isEvery value $e^{-\lambda }{\lambda }^n/n!$ is nonnegative, and their sum is $e^{-\lambda }{e}^{\lambda }=1$. Hence this is a valid probability mass function.

For $n\ge 2$, $X_2+\cdots +X_n$ is independent of $X_1$ and has a Poisson distribution with parameter $(n-1)\lambda$. Part (ii) therefore givesIts probability at zero isFor $n=1$ the event being conditioned on is $X_1=1$, so the answer is $0$, also matching the displayed expression. The cancellation of $\lambda$ reflects the equal rates of all $n$ counts, not an approximation or large-$n$ limit.

For $0\le k\le n$, the event $\{X=k,X+Y=n\}$ is $\{X=k,Y=n-k\}$. Use independence in its numerator and the Poisson distribution of the sum in its denominator:Outside $0\le k\le n$ the conditional probability is zero. ConsequentlyThis binomial distribution includes $n=0$, when it is concentrated at zero. Conditioning on the total converts the independent counts into an allocation of that total between the two rates.

Condition first on $X$. A Poisson distribution with parameter $k$ has conditional probability generating function $e^{k(s-1)}$, including the degenerate zero case when $k=0$. The law of total expectation givesFor $0\le s\le 1$ the inner argument is in $(0,1]$, so this use of the generating function is justified directly. The result is a compound Poisson distribution with a Poisson distribution of unit parameter for each jump, also called a Neyman Type A distribution in this case. It is generally not an ordinary Poisson distribution.

Independence gives the probability generating function of the sum as the product:This is the probability generating function of the Poisson distribution with parameter $\mu +\lambda$. Uniqueness of the coefficients of a power series therefore gives $X+Y\sim \mathrm{Poisson}(\mu +\lambda )$. The product rule for these generating functions requires the stated independence.

For a nonnegative integer-valued random variable, its probability generating function is $G_X(s)=\mathbb{E}[s^X]={\sum }_{k\ge 0}\mathbb{P}(X=k)s^k$, at least for $|s|\le 1$. Insert the Poisson distribution masses and sum the exponential series:

The conditional mean of a Poisson distribution equals its parameter, so $\mathbb{E}[Z\mid X]=X$. The law of total expectation givesAs a check, differentiating the probability generating function in part (iii) at one also gives $G_Z^{\prime }(1)=\mu$.

Subtract the first three values of the probability mass function of the Poisson distribution from one:

For a Poisson distribution with parameter $\lambda >0$, the probability mass function isand is zero at all other values. Its normalization follows from the exponential series ${\sum }_{k\ge 0}{\lambda }^k/k!=e^{\lambda }$.

By the preceding convolution, $Z=X_1+X_2$ has a Poisson distribution with parameter $\nu =\lambda +\mu$. Directly from its probability mass function, shifting the index in the absolutely convergent sums gives $\mathbb{E}[Z]=\nu$ and $\mathbb{E}[Z(Z-1)]={\nu }^2$. Since $Z^2=Z(Z-1)+Z$, $\mathbb{E}[(X_1+X_2{)}^2]=(\lambda +\mu {)}^2+(\lambda +\mu )$. This is also the identity $\mathbb{E}[Z^2]=\mathrm{Var}(Z)+(\mathbb{E}[Z]{)}^2$.

This is the ordinary factorial of $X$, rather than a falling-factorial moment. Substituting the probability mass function of the Poisson distribution gives the nonnegative seriesThe geometric series converges exactly when $0<\lambda <1$. Therefore $\mathbb{E}[X!]=\frac{e^{-\lambda }}{1-\lambda }$ for $0<\lambda <1$. For $\lambda \ge 1$, its terms fail to tend to zero and the expectation is infinite. This is the factorial of a Poisson random variable.

The independent random variables have a product joint probability mass function. For integers $r,t\ge 0$,It is zero for all other pairs. The factors are the respective Poisson distribution masses, with the product justified by independence.

Conditioning on $Z=z$ has positive probability exactly when $z=2r+1$ for an integer $r\ge 0$. This is the same event as $X+Y=r$. The addition of independent Poisson random variables gives a Poisson distribution with parameter $\lambda +\mu$ for the sum. Therefore, for $0\le t\le r$,Thus $X\mid \{Z=2r+1\}\sim \mathrm{Binomial}(r,\lambda /(\lambda +\mu ))$. The case $r=0$ is a constant zero random variable. For even, nonpositive, or noninteger $z$, the conditioning event has zero probability, so an event-conditional probability distribution is not specified by the quotient formula. This is a two-component case of Poisson-multinomial conditioning.

Insert the Poisson distribution probability generating function into the preceding bound:For fixed $x>\lambda$, differentiate the exponent with respect to $s$. Its derivative is $\lambda -x/s$, and its second derivative is $x/s^2>0$, so its unique minimum occurs at $s=x/\lambda >1$. At that value,ConsequentlyThis Chernoff bound remains valid for noninteger positive thresholds $x>\lambda$; the event is then the same as exceeding or equalling its integer ceiling.

Summing the two nonzero entries with $X=1$ givesThe remaining probability is $\mathbb{P}(X=0)=1-p$, so $X$ has the Bernoulli distribution with parameter $p$.

For the other marginal, the two entries at $Y=0$ add to $e^{-p}$, the entry at $Y=1$ is $p{e}^{-p}$, and for $k⩾2$ the only nonzero entry is already $e^{-p}{p}^k/k!$. Thusand $Y$ has the Poisson distribution with parameter $p$. This also checks normalization of the joint probability mass function by summing the marginal.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.