Exponential likelihood + Exponential prior =?

Following Kevin Murphy's MLAPP and I am having troubles with Exercise 3.11. The exercise happens in parts, and I am hoping to get help with multiple parts.

A lifetime $$X$$ of a machine is modeled by an exponential distribution with unknown parameter $$\theta$$. The likelihood is $$p(x \mid \theta) = \theta e^{-\theta x}$$

Assume an expert believes $$\theta$$ should have a prior distribution that is also exponential $$p(\theta) = \text{Expon}(\theta \mid \lambda) = \text{Gamma}(\theta \mid 1, \lambda) = \lambda e^{-\lambda \theta}$$

Note: $$\mathcal{D} = \{ x_1, \dots, x_N \}$$ is the data.

1. What is the posterior $$p(\theta \mid \mathcal{D}, \lambda)$$?

I get \begin{align} p(\theta \mid \mathcal{D}, \lambda) &= p(\theta \mid \lambda) p(\mathcal{D} \mid \theta) \\ &= \lambda e^{-\lambda \theta} \prod_{i=1}^N \theta e^{-\theta x_i} \\ &= \lambda \theta^N e^{-\theta (\lambda + \sum_{i=1}^N x_i)} \end{align}

which doesn't look like any distribution that I know of.

2. Is the exponential prior conjugate to the exponential likelihood?

No. The posterior is not an exponential. But I'm not sure if I computed the posterior correctly.

I assume that I'm doing something wrong, because I'm later asked about the posterior mean, which is difficult to do by integration (integration by parts and extracting a term that depends upon the Gamma function).

• I might not use your notation because I think of the prior as a Gamma distribution rather than an exponential distribution (in your example with shape parameter $1$ and rate parameter $\lambda$) and I might prefer $\pi$ for the density of $\theta$ rather than $p$, but essentially yes. May 7, 2020 at 17:43