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.
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%):
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