Now, suppose that we instead have the proper prior \pi (\lambda ) \sim \textsf{Exp}(a) (a > 0). Again, just as in part (b): conditional on \lambda, we have observations X _1, X _2, \cdots, X _{n} \stackrel{\text {i.i.d}}{\sim } \textsf{N}(\lambda , 1). You may assume that a < \displaystyle \sum _{i=1}^{n} X_ i. Compute the posterior distribution \pi (\lambda | X_1, X_2, \ldots , X_ n), then provide the following statistics on the posterior distribution. Write Phi for the CDF function \Phi () and PhiInv for its inverse.
Use SumXi for \sum _{i=1}^ n X_ i.
median:
unanswered
[Math Processing Error]
mode:
To compute the posterior distribution, we use Bayes' theorem:
\(\pi(\lambda|X_1,X_2,...,X_n) \propto \pi(\lambda) \times \prod_{i=1}^{n} f(X_i|\lambda)\)
where \(\pi(\lambda)\) is the prior distribution and \(f(X_i|\lambda)\) is the likelihood function.
In this case, the prior distribution \(\pi(\lambda)\) is an exponential distribution, given by \(\pi(\lambda) \sim \textsf{Exp}(a)\).
The likelihood function is given by \(f(X_i|\lambda) = \frac{1}{\sqrt{2\pi}}e^{-\frac{(X_i-\lambda)^2}{2}}\).
Therefore, the posterior distribution is given by:
\(\pi(\lambda|X_1,X_2,...,X_n) \propto e^{-a\lambda} \times e^{-\frac{1}{2}\sum_{i=1}^{n}(X_i-\lambda)^2}\)
Multiplying the exponential terms and simplifying, we get:
\(\pi(\lambda|X_1,X_2,...,X_n) \propto e^{-(a+\frac{n}{2})\lambda + \sum_{i=1}^{n}X_i\lambda - \frac{1}{2}\sum_{i=1}^{n}X_i^2}\)
We recognize the exponential distribution in the exponent, so we can factor out the normalizing constant to find the proportional form of the posterior distribution:
\(\pi(\lambda|X_1,X_2,...,X_n) \propto e^{-(a+\frac{n}{2})\lambda}\times e^{-\frac{1}{2}(\sum_{i=1}^{n}X_i^2 - 2\lambda\sum_{i=1}^{n}X_i)}\)
This is a proportional form of a gamma distribution with shape parameter \(a^* = a + \frac{n}{2}\) and rate parameter \(b^* = \sum_{i=1}^{n}X_i\).
Therefore, the posterior distribution is given by \(\pi(\lambda|X_1,X_2,...,X_n) \sim \textsf{Gamma}(a^*, b^*)\).
Now, let's move on to computing the median and mode of the posterior distribution.
The median of a gamma distribution is given by \(\lambda = \frac{a^* - 1}{b^*}\). Therefore, the median of the posterior distribution is:
\(\text{median} = \frac{a + \frac{n}{2} - 1}{\sum_{i=1}^{n}X_i}\)
The mode of a gamma distribution is given by \(\lambda = \frac{a^* - 1}{b^*}\) if \(a^* > 1\), otherwise the mode is undefined. Therefore, the mode of the posterior distribution is:
\(\text{mode} = \frac{a + \frac{n}{2} - 1}{\sum_{i=1}^{n}X_i}\) if \(a + \frac{n}{2} > 1\), otherwise the mode is undefined.
To compute the posterior distribution \(\pi(\lambda | X_1, X_2, \ldots, X_n)\), we need to use Bayes' theorem. The posterior distribution is given by:
\[
\pi(\lambda | X_1, X_2, \ldots, X_n) \propto f(X_1, X_2, \ldots, X_n | \lambda) \cdot \pi(\lambda)
\]
where \(f(X_1, X_2, \ldots, X_n | \lambda)\) is the likelihood function of the data given the parameter \(\lambda\), and \(\pi(\lambda)\) is the prior distribution.
In this case, the likelihood function \(f(X_1, X_2, \ldots, X_n | \lambda)\) is the product of the densities of the normal distributions:
\[
f(X_1, X_2, \ldots, X_n | \lambda) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(X_i - \lambda)^2}{2}\right)
\]
And the prior distribution is given by \(\pi(\lambda) \sim \textsf{Exp}(a)\), which means it follows an exponential distribution with rate parameter \(a\).
To simplify the calculation, we can work with the log posterior distribution instead:
\[
\log \left(\pi(\lambda | X_1, X_2, \ldots, X_n)\right) \propto \sum_{i=1}^{n} \left(-\frac{(X_i - \lambda)^2}{2}\right) - a\lambda
\]
Now we can find the mode of the posterior distribution by maximizing this expression. Taking the derivative with respect to \(\lambda\) and setting it equal to zero, we get:
\[
\sum_{i=1}^{n} (X_i - \lambda) - a = 0
\]
Simplifying further, we find:
\[
\lambda = \frac{\sum_{i=1}^{n} X_i + a}{n}
\]
This gives us the mode of the posterior distribution.
To compute the median, we need to find the value of \(\lambda\) such that the cumulative distribution function (CDF) of the posterior distribution is equal to 0.5. This can be done using numerical methods or by finding the inverse of the CDF (if available).
Unfortunately, the exact form of the posterior distribution is not provided, so it is not possible to provide a step-by-step calculation for the median or other statistics without additional information.
If you provide the specific values of \(a\) and the observed data \(X_1, X_2, \ldots, X_n\), I can help you with the calculations.