Let $X$ be a random sample from an exponential distribution with mean $1/\lambda$. If $\lambda$ has a prior distribution with probability density function $g(\lambda) = \begin{cases} \lambda e^{-\lambda} & ; \quad \lambda > 0 \\ 0 & ; \quad \lambda \leq 0 \end{cases}$ then the Bayes estimator of $1/\lambda$ with respect to the squared error loss function is
The problem asks for the Bayes estimator of $1/\lambda$ for an exponential distribution, given a specific prior distribution for $\lambda$ and a squared error loss function. We assume $X$ represents a single observation from the distribution.
Likelihood and Prior
The probability density function (PDF) of the exponential distribution with parameter $\lambda$ is $f(x|\lambda) = \lambda e^{-\lambda x}$ for $x > 0$. The mean is $1/\lambda$. The likelihood function given an observation $X=x$ is:
$ L(\lambda|x) = \lambda e^{-\lambda x} $
The prior distribution for $\lambda$ is given by:
$ g(\lambda) = \lambda e^{-\lambda} \quad \text{for } \lambda > 0 $
Posterior Distribution
The posterior distribution PDF, $p(\lambda|x)$, is proportional to the product of the likelihood and the prior:
$ p(\lambda|x) \propto L(\lambda|x) g(\lambda) $
$ p(\lambda|x) \propto (\lambda e^{-\lambda x}) (\lambda e^{-\lambda}) $
$ p(\lambda|x) \propto \lambda^2 e^{-\lambda x - \lambda} $
$ p(\lambda|x) \propto \lambda^2 e^{-\lambda(x+1)} $
This form is proportional to the kernel of a Gamma distribution, $\lambda^{k-1} e^{-c\lambda}$. By comparing the exponents, we identify the shape parameter $k=3$ and the rate parameter $c = x+1$.
Therefore, the posterior distribution for $\lambda$ is:
$ \lambda | X=x \sim \text{Gamma}(k=3, c=x+1) $
The full posterior PDF is $p(\lambda|x) = \frac{(x+1)^3}{\Gamma(3)} \lambda^{2} e^{-\lambda(x+1)}$.
Bayes Estimator Calculation
For a squared error loss function, the Bayes estimator of a parameter is the posterior mean of that parameter.
We need to find the posterior mean of $1/\lambda$:
$ \hat{\theta}_{SE}(x) = E[1/\lambda | X=x] $
For a random variable $Y \sim \text{Gamma}(k, c)$, the mean of its reciprocal, $E[1/Y]$, is given by $\frac{c}{k-1}$, provided $k > 1$.
In our case, $Y = \lambda$, $k=3$, and $c=x+1$. Since $k=3 > 1$, we can use the formula:
$ E[1/\lambda | X=x] = \frac{c}{k-1} = \frac{x+1}{3-1} $
$ E[1/\lambda | X=x] = \frac{x+1}{2} $
Conclusion
The Bayes estimator of $1/\lambda$ with respect to the squared error loss function is $\frac{x+1}{2}$.
Important Questions from Elementary Bayesian Inference
- Suppose the distribution of $X$ given $\theta$ is normal with mean $\theta$ and variance $15$. Further, let the prior (improper) distribution of $\theta$ be proportional to $1, \ -\infty<\theta<\infty$. If the observed value of $X$ is $13$, then which of the following statements is true?
- Let $X_1, X_2, . . ., X_n$ be a random sample from $N(\theta, 1)$, $\theta \in R$. If $\hat{\theta}$ is the Bayes estimator of $\theta$ with respect to some prior $\pi(\theta)$ and loss function $L(\theta, d)$. Then, which of the following statements are true?
- Let $X|\theta \sim \text{Uniform}[0, \theta]$ and $\theta$ has an Exponential distribution with mean $\lambda$, where $\lambda > 3$ is known. If the realized value of $X$ is $2025$, then the posterior mode equals
- $X_1, X_2, \cdots, X_n$ are independent and identically distributed $N(\theta, 1)$ random variables, where $\theta$ takes only integer values i.e.
$\theta \in \{\cdots, -2, -1, 0, 1, 2, \cdots\}$.
Which of the following is the maximum likelihood estimator of $\theta$? - Suppose the probability mass function of a random variable X under the parameter $\theta = \theta_0$ and $\theta = \theta_1 (\ne \theta_0)$ are given by
x 0 1 2 3 $p_{\theta_0}(x)$ 0.01 0.04 0.5 0.45 $p_{\theta_1}(x)$ 0.02 0.08 0.4 0.5
Define a test $\phi$ such that $\phi(x) = 1$ if $x = 0, 1$, and $0$ if $x = 2, 3$.
For testing $H_0: \theta = \theta_0$ against $H_1: \theta = \theta_1$, the test $\phi$ is
