Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with parameter \(\theta, 0<\theta<\infty .\) Let \(Y=\sum_{1}^{n} X_{i}\) and let \(\mathcal{L}[\theta, \delta(y)]=[\theta-\delta(y)]^{2} .\) If we restrict our considerations to decision functions of the form \(\delta(y)=b+y / n\), where \(b\) does not depend on \(y\), show that \(R(\theta, \delta)=b^{2}+\theta / n .\) What decision function of this form yields a uniformly smaller risk than every other decision function of this form? With this solution, say \(\delta\), and \(0<\theta<\infty\), determine \(\max _{\theta} R(\theta, \delta)\) if it exists.

Short answer

The decision function that yields the smallest risk is \( \delta(y) = y/n \). The risk function does not have a maximal value over the given domain \( 0<\theta<\infty \) as the risk increases along with \( \theta \).

Step-by-step solution

Compute the Risk Function

The risk function is given by the expected value of the loss function \( \mathcal{L}[\theta, \delta(y)] \), thus it is defined as: \( R(\theta, \delta) = E[\mathcal{L}[\theta, \delta(y)]] \). Substitute the given loss function, \( \mathcal{L}[\theta, \delta(y)] = [\theta - \delta(y)]^2 \) and decision function, \( \delta(y) = b + y/n \) into the risk function, it becomes: \( R(\theta, \delta) = E[(\theta - (b + y/n))^2] \).

Simplify the Risk Function

According to the properties of expectation \( E[X + Y] = E[X] + E[Y] \) and \( E[cX] = cE[X] \) where \( X \) and \( Y \) are two random variables and \( c \) is a constant, the risk function can be rewritten as: \( R(\theta, \delta) = E[\theta^2 - 2\theta(b + y/n) + (b + y/n)^2] \).

Calculate the Expectation

The expectation of the sum of independent Poisson random variables is the sum of their individual expectations. Since \( X_i \) follows a Poisson distribution with parameter \( \theta \), the expectation is \( E[X_i] = \theta \), then \( E[Y] = E[\sum_{1}^{n} X_i] = n\theta \) which simplifies the Risk Function to \( R(\theta, \delta) = b^2 + \theta/n \).

Find the Minimum Risk

It is required to find a decision function that minimizes the risk. To do that, take a derivative of the risk function with respect to \( b \), set it equal to zero and solve for \( b \). The derivative of the function \( b^2 + \theta/n \) with respect to \( b \) is \( 2b = 0 \), which gives \( b = 0 \). Thus the decision function that minimizes the risk is \( \delta(y) = y/n \).

Determine the Maximum Risk

With \( \delta = y/n \), the risk function \( R(\theta, \delta) = \theta/n \). This does not have a maximum value over the given domain \( 0<\theta<\infty \), since as \( \theta \) increases, \( R(\theta, \delta) \) also increases.