Numerical values of parameters of posterior distribution

Question

Question

Investigators are wishing to perform Bayesian analysis for drug treatment duration that can be described using a exponential distribution with paramater $\lambda$ describing drug duration in months.

\begin{align*} f_X(x) = \lambda e^{-\lambda x} \quad x \geq 0, \; \lambda > 0 \end{align*}

With expected value

\begin{align*} \mathbb{E}(X) = \frac{1}{\lambda} \end{align*}

and cumulative distribution function

\begin{align*} F_X(x) = \begin{cases} 1 - e^{-\lambda x} &\; x \geq 0 \\ 0 & \; x < 0 \end{cases} \end{align*} Previously it had been found the median duration of drug was 3 months.

The investigators have chosen a Gamma distribution as a prior distribution of parameter $\lambda$

\begin{align*} f(\lambda; \alpha, \beta) = \frac{\beta^{\alpha}\lambda^{\alpha - 1}e^{-\beta \lambda}}{\Gamma(\alpha)} \; \lambda >0 \; \alpha > 0 \; \beta > 0 \end{align*}

With expected value

\begin{align*} \mathbb{E}(X) = \frac{\alpha}{\beta} \end{align*}

The investigators perform a study in n = 17 participants using a new formulation of the drug that they hope will enable longer duration of treatment than the current standard. These participants have a mean duration of 4.9 months. We assume that the prior value of $\alpha = 3$.

Given the numerical values of the parameters of the posterior distribution

My attempt

Using the CDF, the median value of $x$ can be shown to be $\frac{\ln(2)}{\lambda}$. Using the fact that the median time in months was 3. Therefore, the expected value of $\lambda$ in the would be $\lambda = \frac{\ln2}{3}$

Calculating the likelihood function for the exponential distribution.

\begin{align} l(x_i | \lambda) = \lambda^{n}e^{-\lambda\sum_{i =1}^{n} x_i} \end{align}

Then to perform the Bayesian analysis, we can examine the proportionality of the numerator with regards to the likelihood function and the prior

\begin{align} \mathrm{Pr}(\lambda | x_i) \propto \mathrm{Pr}(x_i | \lambda) \cdot \mathrm{Pr}(\lambda) \end{align}

This results in the following posterior distribution with omission of terms that do not depend on parameter $\lambda$.

\begin{align} \mathrm{Pr}(\lambda | x_i) \propto &= \lambda^{n+\alpha - 1}e^{-\lambda\left(\sum_{i = 1}^{n} x_i + \beta \right)} \end{align}

Therefore, the posterior distribution follows a shape of a Gamma distribution.

\begin{align*} \mathrm{Pr}(\lambda | x_i) \sim \mathrm{Gamma}\left(\alpha + n - 1, \beta + \sum_{i = 1}^{n} x_i\right) \end{align*}

Concerns

I am unsure where to go with this to give numerical estimates for each parameter. Would I have to use the fact that I know the number of $n$ participants now to obtain $\sum x_i$ by the mean? Any help would be appreciated.

Thank you!

Answer 1

I am not sure where the $-1$ comes from in $\alpha + n-1$ if your original prior is $\mathrm{Gamma}\left(\alpha, \beta \right)$. I would have thought a posterior $\mathrm{Gamma}\left(\alpha+n, \beta +\sum x_i\right) = \mathrm{Gamma}\left(\alpha+n, \beta +n\bar x\right) $ was what you would want.

You have $\alpha=3$ for your prior; you also need $\beta$. Perhaps you are supposed to deduce it from "Previously it had been found the median duration of drug was $3$ months." There does not seem to be an obviously correct way of doing this but you could think about using your "the expected value of $\lambda$ in the would be $\lambda = \frac{\ln 2}{3}$" and then divide $\alpha=3$ by this to suggest using $\beta\approx 12.984$.

So I would be tempted to use a $\mathrm{Gamma}\left(\alpha+n, \beta +n\bar x\right)$ posterior for the rate $\lambda$ where $\alpha+n = 3+17 = 20$ and $\beta +n\bar x = 12.984 +17 \times 4.7 \approx 92.884$.