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

Probability Theory Solution – Problem 66: Determine all possible distributions of

Question

Let $X$ be a random variable. 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

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

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

1. Use the independence property Since $X$ and $Y$ are independent, the probability generating function of their sum $Z = X + Y$ equals the product of their individual probability generating functions: $G_{Z}(s) = G_{X}(s) \cdot G_{Y}(s)$

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

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

2. Discuss the validity of the parameter For $X$ to be a valid random variable (with nonnegative probabilities), the corresponding Poisson parameter must be nonnegative. That is, we require $\mu - \lambda \geqslant 0$ , or equivalently $\mu \geqslant \lambda$ . * 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 cannot arise under the premise that the sum of independent variables yields a Poisson distribution (i.e., the parameter of $X+Y$ must be at least as large as 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., $P(X=0)=1$ , meaning $X$ is the constant 0. * Case 3: If $\mu > \lambda$ , then $X \sim P(\mu - \lambda)$ , a standard Poisson distribution.

3. 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 is the marking rubric 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 (probability generating function / characteristic function / moment generating function) (recommended approach)

Establish the transform relation (2 pts) [additive]:
Introduce an appropriate transform tool (e.g., probability generating function $G(s)$ , characteristic function $\phi(t)$ , or moment generating function $M(t)$ ).
And use the independence of $X, Y$ to explicitly write the product relation (e.g., $G_{X+Y}(s) = G_X(s) \cdot G_Y(s)$ ).
*If only the definition is listed 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 expression 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 function form, explicitly state that $X$ follows a Poisson distribution with parameter $\mu - \lambda$ .
*If only the transform expression is retained without naming the distribution, deduct 1 pt.*
Discussion of parameter validity (1 pt) [additive]:
State that the parameter must be nonnegative ( $\mu \ge \lambda$ 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 checkpoint.*

Chain B: Using Raikov's Theorem (and cumulant / characteristic number analysis)

Cite the decomposition theorem (4 pts) [additive]:
Explicitly cite 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 component must follow a Poisson distribution."
*Note: This is the key theoretical basis for establishing the existence of the distribution form; the theorem name or 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 accurately derive that the parameter of $X$ is $\mu - \lambda$ .
Discussion of parameter validity (1 pt) [additive]:
State that the parameter must be nonnegative ( $\mu \ge \lambda$ ).

Total (max 7)

2. Zero-credit items

Merely copying the given conditions from the problem (e.g., $X \perp Y$ , $Y \sim P(\lambda)$ ).
Only listing the Poisson probability formula ( $P(X=k)=\dots$ ) without any concrete steps toward deriving the distribution of $X$ .
Guessing 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 merely uses the property that "the sum of two Poisson distributions is a Poisson distribution" (a sufficient condition) to directly assert "therefore $X$ must be Poisson" (necessity), without using the transform method to prove uniqueness or citing Raikov's theorem. Such solutions are considered logically incomplete; award at most 3 pts (for parameter computation and conclusion portions).
Missing parameter range (Flat -1):
Derives $X \sim P(\mu-\lambda)$ but does not mention the physical constraint $\mu \ge \lambda$ .
Symbol confusion (Flat -1):
Confuses random variables (uppercase $X$ ) with values or parameters during the derivation, causing unclear logical exposition.

Probability Theory – Problem 66: Determine all possible distributions of