Let's break this down clearly, starting from the given unbiasedness condition and working our way to the sum.

1. Simplify the Unbiasedness Equation #

We know $\delta(X)$ is an unbiased estimator of $g(\lambda) = e^{{-\lambda}(3\lambda}2 + 2\lambda + 1)$, so:
$$E(\delta(X)) = \sum_{k=0}^\infty \frac{\delta(k)e^{{-\lambda}\lambda}k}{k!} = e^{{-\lambda}(3\lambda}2 + 2\lambda + 1)$$

Since $e^{-\lambda} \neq 0$ for all $\lambda > 0$, we can divide both sides by $e^{-\lambda}$ to eliminate this common factor:
$$\sum_{k=0}^\infty \frac{\delta(k)\lambda^k}{k!} = 3\lambda^2 + 2\lambda + 1$$

2. Equate Power Series Coefficients #

The left-hand side is a power series in $\lambda$, and the right-hand side is a polynomial of degree 2. Power series have unique coefficients, so we can match the coefficients of each $\lambda^k$ term on both sides:

• For $k=0$ (constant term): $\frac{\delta(0)}{0!} = 1$ → $\delta(0) = 1 \times 0! = 1$
• For $k=1$ (coefficient of $\lambda$): $\frac{\delta(1)}{1!} = 2$ → $\delta(1) = 2 \times 1! = 2$
• For $k=2$ (coefficient of $\lambda^2$): $\frac{\delta(2)}{2!} = 3$ → $\delta(2) = 3 \times 2! = 6$
• For $k \geq 3$: The polynomial has no terms with $\lambda^k$, so $\frac{\delta(k)}{k!} = 0$ → $\delta(k) = 0$

3. Compute the Sum #

Now we just add up all non-zero $\delta(k)$ values (since $\delta(k)=0$ for $k \geq 3$, they contribute nothing to the sum):
$$\sum_{k=0}^\infty \delta(k) = \delta(0) + \delta(1) + \delta(2) + 0 + 0 + \dots = 1 + 2 + 6 = 9$$

Verification #

To confirm this is correct, check that $\delta(X)$ is indeed unbiased:
$$E(\delta(X)) = 1 \cdot e^{-\lambda} + 2 \cdot \lambda e^{-\lambda} + 6 \cdot \frac{\lambda^2 e^{-\lambda}}{2!} = e^{-\lambda}(1 + 2\lambda + 3\lambda^2)$$
Which matches the target function $g(\lambda) = e^{{-\lambda}(3\lambda}2 + 2\lambda + 1)$.

内容的提问来源于stack exchange，提问作者Vishavjeet Singh