Show posterior mean can be written as a weighted average of the prior mean and MLE

Question

Suppose $Y_1, \dots Y_n$ are exponentially distributed: $Y_i | \lambda \sim Exp(\lambda)$. Find the conjugate prior for $\lambda$, and the corresponding posterior distribution. Show that the posterior mean for the failure rate $\lambda$ can be written as a weighted average of the prior mean and the maximum likelihood estimator, $\hat{\lambda}=\bar{y}^{-1}$.

Given that the joint pdf as follow:-

$P(Y_i | \lambda)=\lambda^{n}e^{-\lambda \sum_{i=1}^{n} Y_i}$

Then the likelihood function is

$ L (\lambda | Y_i)\propto \lambda^{n}e^{-\lambda \sum_{i=1}^{n}Y_i}$

which give Gamma distribution with $\alpha = n+1$ and $\beta =\sum_{i=1}^{n} Y_i$.

Thus, with the prior density

$P(\lambda)\propto \lambda^{a-1} e^{- \lambda b}$

gives the posterior distribution as follow:-

$P(\lambda | Y_i) \propto \lambda^{n+a-1} e^{-\lambda (\sum_{i=1}^{n}Y_i + b)}$

and resulting in Gamma distribution with $\alpha = n+a$ and $\beta = \sum_{i=1}^{n} Y_i + b$.

My problem now is how can I show the posterior mean for the failure rate λ can be written as a weighted average of the prior mean and the maximum likelihood estimator?

As far I get as the following:-

$E[\lambda | Y] = \frac{\alpha}{\beta} = \frac{n+a}{\sum_{i=1}^{n} Y_i + b}$

Can anyone help me to show how the posterior mean above can be written as a weighted average of the prior mean and the maximum likelihood estimator, $\hat{\lambda}=\bar{y}^{-1}$?

Answer 1

Continuing from where you left, $$\begin{align} \bar\lambda_{post}^{-1} &= \frac{\sum_{i}y_i+b}{n+a} \\ &= \frac{n}{n+a}\frac{\sum_{i}y_i}{n} + \frac{a}{n+a}\frac{b}{a} \\ &= \rho\lambda_{mle}^{-1} + (1-\rho)\bar\lambda_{prior}^{-1} \\ \end{align}$$

where $\rho = \frac{n}{n+a}$ and $\lambda_{mle}$, $\bar\lambda_{prior}$ and $\bar\lambda_{post}$ denote the maximum likelihood estimator, prior mean and posterior mean respectively.