Distribution Analysis

We are given a joint discrete distribution concerning random variables $(X,Y)$. The properties provided are:

• The conditional distribution of $X$ given $Y=y$ follows a Binomial distribution: $X \mid Y=y \sim \text{Binomial} (y, 0.5)$.
• The marginal distribution of $Y$ follows a Poisson distribution: $Y \sim \text{Poisson}(\lambda)$, where $\lambda > 0$.
• $T = T(X,Y)$ represents any unbiased estimator of the parameter $\lambda$. An estimator $T$ is unbiased if its expected value equals the true parameter value, i.e., $E[T] = \lambda$.

Variance of Y

For a random variable $Y$ following a Poisson distribution with parameter $\lambda$, denoted as $Y \sim \text{Poisson}(\lambda)$, both its mean and variance are equal to $\lambda$.

Therefore, we have $Var(Y) = \lambda$.

Variance Bound for Unbiased Estimators

The question asks about the variance of an arbitrary unbiased estimator $T$ of $\lambda$. We can use the Cramér-Rao Lower Bound (CRLB) to establish a minimum possible variance for such estimators.

The CRLB depends on the Fisher Information, $I(\lambda)$, derived from the distribution of $Y$, as $Y$ is directly related to $\lambda$.

The probability mass function (PMF) of $Y \sim \text{Poisson}(\lambda)$ is $p(y; \lambda) = \frac{e^{-\lambda}\lambda^y}{y!}$ for $y = 0, 1, 2, \dots$.

The natural logarithm of the PMF (log-likelihood) is $\ln p(y; \lambda) = -\lambda + y \ln \lambda - \ln(y!)$.

To find the Fisher Information, we first compute the second derivative of the log-likelihood with respect to $\lambda$:

$ \frac{\partial^2}{\partial \lambda^2} \ln p(y; \lambda) = \frac{\partial}{\partial \lambda} \left( -1 + \frac{y}{\lambda} \right) = -\frac{y}{\lambda^2} $

The Fisher Information $I(\lambda)$ is the negative expectation of this second derivative:

$ I(\lambda) = -E\left[\frac{\partial^2}{\partial \lambda^2} \ln p(Y; \lambda)\right] = -E\left[-\frac{Y}{\lambda^2}\right] = \frac{1}{\lambda^2} E[Y] $

Since $E[Y] = \lambda$ for the Poisson distribution, we get:

$ I(\lambda) = \frac{1}{\lambda^2} (\lambda) = \frac{1}{\lambda} $

The CRLB states that the variance of any unbiased estimator $T$ of $\lambda$ must be greater than or equal to $1/I(\lambda)$:

$ Var(T) \ge \frac{1}{I(\lambda)} $

Substituting the calculated $I(\lambda)$:

$ Var(T) \ge \frac{1}{1/\lambda} $

$ Var(T) \ge \lambda $

Conclusion

From the CRLB, we have shown that $Var(T) \ge \lambda$ must hold true for any unbiased estimator $T$ of $\lambda$.

We also established earlier that $Var(Y) = \lambda$, because $Y \sim \text{Poisson}(\lambda)$.

Therefore, the inequality $Var(T) \ge \lambda$ is equivalent to $Var(T) \ge Var(Y)$.

This confirms that both statements are correct:

• $Var(T) \ge Var(Y)$ for all $\lambda$.
• $Var(T) \ge \lambda$ for all $\lambda$.