Let be a random variable. Suppose there exists a random variable , independent of , such that both and follow Poisson distributions. Determine all possible distributions of .

Probability Theory Solution – Problem 15: Determine all possible distributions of

Question

Let $X$ be a random variable. Suppose there exists a random variable $Y$ , independent of $X$ , such that both $Y$ and $X+Y$ follow Poisson distributions. Determine all possible distributions of $X$ .

Step-by-step solution

Step 1. Define the probability generating function. For a random variable $\xi$ , define its probability generating function (PGF) as $G_{\xi}(s) = E[s^{\xi}] = \sum_{k=0}^{\infty} P(\xi=k)s^k$ . For a Poisson distribution with parameter $\lambda$ , the PGF is: $G(s) = e^{\lambda(s-1)}$

Step 2. Set up the parameters and express the PGFs of $Y$ and $X+Y$ . Let $Y \sim P(\lambda)$ with $\lambda > 0$ . Then: $G_{Y}(s) = e^{\lambda(s-1)}$ Let $Z = X + Y$ with $Z \sim P(\mu)$ and $\mu > 0$ . Then: $G_{Z}(s) = e^{\mu(s-1)}$

Step 3. Apply the independence property. Since $X$ and $Y$ are independent, the PGF of their sum $Z = X + Y$ equals the product of their individual PGFs: $G_{Z}(s) = G_{X}(s) \cdot G_{Y}(s)$

Step 4. Solve for $G_{X}(s)$ . Substituting the known expressions into the equation above: $e^{\mu(s-1)} = G_{X}(s) \cdot e^{\lambda(s-1)}$ Solving yields: $G_{X}(s) = \frac{e^{\mu(s-1)}}{e^{\lambda(s-1)}} = e^{\mu(s-1) - \lambda(s-1)} = e^{(\mu - \lambda)(s-1)}$

Step 5. Identify the distribution. Observe that $G_{X}(s) = e^{(\mu - \lambda)(s-1)}$ is precisely the PGF of a Poisson distribution with parameter $\theta = \mu - \lambda$ . This implies that $X$ must follow a Poisson distribution with parameter $\mu - \lambda$ .

Step 6. Discuss the validity of the parameter. For $X$ to be a well-defined random variable (with non-negative probabilities), the corresponding Poisson parameter must be non-negative. That is, we require $\mu - \lambda \geqslant 0$ , i.e., $\mu \geqslant \lambda$ . * Case 1: If $\mu < \lambda$ , then the resulting $P(X=k)$ would alternate in sign or become complex, which does not constitute a valid probability distribution. Hence this case is impossible under the hypothesis that $X+Y$ is Poisson with $X$ and $Y$ independent (the parameter of $X+Y$ must be at least that of $Y$ ). * Case 2: If $\mu = \lambda$ , then $G_{X}(s) = e^{0} = 1$ . This corresponds to the degenerate Poisson distribution (parameter 0), i.e., $P(X=0)=1$ , so $X$ is the constant 0. * Case 3: If $\mu > \lambda$ , then $X \sim P(\mu - \lambda)$ , a standard Poisson distribution.

Step 7. Conclusion. $X$ follows a Poisson distribution $P(\theta)$ with parameter $\theta \geqslant 0$ . Specifically, if $Y \sim P(\lambda)$ and $X+Y \sim P(\mu)$ , then $X \sim P(\mu - \lambda)$ , and the constraint $\mu \geqslant \lambda$ must hold.

Final answer

$X$ follows a Poisson distribution (including the degenerate case with parameter 0).

Marking scheme

The following rubric is based on the official solution approach:

1. Checkpoints (max 7 pts total)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Transform Method (PGF / Characteristic Function / MGF) (Recommended Approach)

Establish the transform relation (2 pts) [additive]:
Introduce an appropriate transform tool (e.g., PGF $G(s)$ , characteristic function $\phi(t)$ , or MGF $M(t)$ ).
And invoke the independence of $X$ and $Y$ to write down the product relation explicitly (e.g., $G_{X+Y}(s) = G_X(s) \cdot G_Y(s)$ ).
*If only the definition is stated without establishing the relation between $X$ and $X+Y$ , award 0 pts.*
Compute the transform of $X$ (2 pts) [additive]:
Substitute the specific Poisson transform formula (e.g., $e^{\lambda(s-1)}$ or $e^{\lambda(e^{it}-1)}$ ).
Use algebraic manipulation to correctly solve for the transform of $X$ (e.g., $G_X(s) = \frac{e^{\mu(s-1)}}{e^{\lambda(s-1)}} = e^{(\mu-\lambda)(s-1)}$ ).
Identify the distribution type (2 pts) [additive]:
Based on the derived functional form, explicitly state that $X$ follows a Poisson distribution with parameter $\mu - \lambda$ .
*If only the transform expression is given without naming the distribution, deduct 1 pt.*
Discussion of parameter validity (1 pt) [additive]:
State that the parameter must be non-negative ( $\mu \ge \lambda$ or $\mu - \lambda \ge 0$ ), or discuss the degenerate case $\mu = \lambda$ where $X$ reduces to the constant 0.
*If the parameter range is not discussed, no credit for this item.*

Chain B: Via Raikov's Theorem (and Cumulant / Characteristic Function Analysis)

Cite the decomposition theorem (4 pts) [additive]:
Explicitly invoke Raikov's theorem (or the equivalent application of Cramer's decomposition theorem to the Poisson distribution), arguing that if the sum of independent random variables follows a Poisson distribution, then each summand must itself follow a Poisson distribution.
*Note: This is the key theoretical justification for the existence of the distributional form; the theorem name or its content must be explicitly mentioned.*
Determine the parameter (2 pts) [additive]:
Use the additivity of expectations, variances, or cumulants (e.g., $E[X+Y] = E[X] + E[Y]$ ) to correctly derive that the parameter of $X$ is $\mu - \lambda$ .
Discussion of parameter validity (1 pt) [additive]:
State that the parameter must be non-negative ( $\mu \ge \lambda$ ).

Total (max 7)

2. Zero-credit items

Merely copying the given conditions from the problem statement (e.g., $X \perp Y$ , $Y \sim P(\lambda)$ ).
Merely listing the Poisson probability formula ( $P(X=k)=\dots$ ) without any concrete steps toward deriving the distribution of $X$ .
Conjecturing that $X$ follows some other distribution (e.g., binomial, normal) and attempting verification, leading to contradictions or computational errors.

3. Deductions

Logical inversion / circular reasoning (Cap at 3/7):
The student uses only the property that the sum of two independent Poisson random variables is Poisson (a sufficient condition) to directly assert that $X$ must be Poisson (necessity), without employing the transform method to prove uniqueness or citing Raikov's theorem. Such a solution is considered logically incomplete and is capped at 3 pts (awarded for parameter computation and conclusion).
Missing parameter range (Flat -1):
Derives $X \sim P(\mu-\lambda)$ but does not mention the constraint $\mu \ge \lambda$ .
Symbol confusion (Flat -1):
Confuses random variables (uppercase $X$ ) with their realizations or parameters during the derivation, resulting in unclear logical exposition.

Probability Theory – Problem 15: Determine all possible distributions of