Suppose we have the improper prior \pi (\lambda ) \propto e^{-a\lambda }, \lambda \in \mathbb {R} (and a \geq 0). Conditional on \lambda, we have observations X _1, X _2, \cdots, X _{n} \stackrel{\text {i.i.d}}{\sim } \textsf{N}(\lambda , 1). Compute the posterior distribution \pi (\lambda | X_1, X_2, \ldots , X_ n), then provide the following statistics on the posterior distribution.
Use SumXi for \sum _{i=1}^ n X_ i.
mean:
variance:
q_{0.025} (cutoff for highest 2.5%):
To compute the posterior distribution, we can use Bayes' theorem:
\pi(\lambda | X_1, X_2, \ldots , X_n) \propto \pi(\lambda) \prod_{i=1}^n f(X_i|\lambda),
where f(X_i | \lambda) is the density of the normal distribution with mean \lambda and variance 1.
Plugging in the given prior and likelihood, we get:
\pi(\lambda | X_1, X_2, \ldots , X_n) \propto e^{-a\lambda} \prod_{i=1}^n \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(X_i - \lambda)^2}.
Taking the logarithm of this expression, we have:
\log(\pi(\lambda | X_1, X_2, \ldots , X_n)) \propto -a\lambda + \sum_{i=1}^n \left(-\frac{1}{2}(X_i - \lambda)^2\right).
To find the posterior distribution, we need to normalize this expression by finding the constant of proportionality. This requires integrating the above expression with respect to \lambda, which leads to a standard normalizing constant involving the square root of a multiple of n.
After normalizing, we find that the posterior distribution is also a normal distribution, with mean and variance given by:
mean: \hat{\lambda} = \frac{\sum_{i=1}^n X_i + a}{n + 1}
variance: \hat{\sigma}^2 = \frac{1}{n + 1}
So, the mean of the posterior distribution is \frac{\sum_{i=1}^n X_i + a}{n + 1} and the variance is \frac{1}{n + 1}.
To compute the cutoff for the highest 2.5% (q_{0.025}), we can use the quantile function of the normal distribution. The quantile for the highest 2.5% is the value q such that P(\lambda > q) = 0.025, which can be computed using a standard statistical software or table for the normal distribution.
To compute the posterior distribution \(\pi(\lambda | X_1, X_2, \ldots, X_n)\), we can use Bayes' theorem:
\[
\pi(\lambda | X_1, X_2, \ldots, X_n) \propto \pi(\lambda) \times \prod_{i=1}^{n} f(X_i|\lambda)
\]
where \(\pi(\lambda)\) is the improper prior distribution and \(f(X_i|\lambda)\) is the likelihood function, which in this case is the probability density function of a normal distribution.
Let's calculate the posterior distribution step-by-step:
Step 1: Compute the product of the likelihoods
\[
\prod_{i=1}^{n} f(X_i|\lambda) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(X_i-\lambda)^2\right)
\]
Expanding this product gives:
\[
\prod_{i=1}^{n} f(X_i|\lambda) = \frac{1}{(2\pi)^{n/2}}\exp\left(-\frac{1}{2}\sum_{i=1}^{n}(X_i-\lambda)^2\right)
\]
Step 2: Apply the prior distribution
\[
\pi(\lambda | X_1, X_2, \ldots, X_n) \propto e^{-a\lambda} \times \frac{1}{(2\pi)^{n/2}}\exp\left(-\frac{1}{2}\sum_{i=1}^{n}(X_i-\lambda)^2\right)
\]
Step 3: Simplify the expression
\[
\pi(\lambda | X_1, X_2, \ldots, X_n) \propto e^{(-a\lambda - \frac{1}{2}\sum_{i=1}^{n}(X_i-\lambda)^2)}
\]
Step 4: Recognize the posterior distribution
The expression in step 3 is proportional to the kernel of a normal distribution. By completing the square, we can rewrite the expression as:
\[
\pi(\lambda | X_1, X_2, \ldots, X_n) \propto \exp\left(-\frac{1}{2}\left(\lambda^2 -2\lambda\left(\frac{a}{n} + \frac{\sum_{i=1}^n X_i}{n}\right) + \text{constant}\right)\right)
\]
This is a normal distribution in \(\lambda\) with mean \(\frac{a}{n} + \frac{\sum_{i=1}^n X_i}{n}\) and variance \(\frac{1}{n}\).
Therefore, the statistics of the posterior distribution are:
Mean: \(\frac{a}{n} + \frac{\sum_{i=1}^n X_i}{n}\)
Variance: \(\frac{1}{n}\)
Cutoff for highest 2.5% (quantile): To compute this, we can use the cumulative distribution function (CDF) of the normal distribution. The cutoff will be the value for which the CDF is equal to 0.975 (since we want the top 2.5%).
Let \(q_{0.025}\) denote the cutoff. We can use the mean and variance of the posterior distribution to find it. Specifically, we can use the quantile function (inverse CDF) of the normal distribution with the mean and variance to find the cutoff value.
Hence:
\(q_{0.025}\) (cutoff for highest 2.5%): calculate as \(\text{{mean}} + z \times \sqrt{\text{{variance}}}\), where \(z\) is the \(z\)-value corresponding to the 0.975 quantile of a standard normal distribution.