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
- 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
- Suppose $X|\theta \sim \text{Binomial}(7,\theta)$, $0 < \theta < 1$, and the prior distribution of $\theta$ is $\text{Beta}(\alpha, \beta)$ where $\alpha > 0$ and $\beta > 0$ are known. Then which of the following statements MAY NOT be true?
Let $X_1, X_2, \dots, X_7$ be a random sample from $N(\mu, \sigma^2)$ where $\mu$ and $\sigma^2$ are unknown. Consider the problem of testing $H_0: \mu = 2$ against $H_1: \mu > 2$. Suppose the observed values of $x_1, x_2, \dots, x_7$ are $1.2, 1.3, 1.7, 1.8, 2.1, 2.3, 2.7$. If we use the Uniformly Most Powerful test, which of the following is true?
Suppose $X_i \mid \theta_i \sim N(\theta_i, \sigma^2), i = 1, 2$ are independently distributed. Under the prior distribution, $\theta_1$ and $\theta_2$ are i.i.d $N(\mu, \tau^2)$, where $\sigma^2, \mu$ and $\tau^2$ are known. Then which of the following is true about the marginal distributions of $X_1$ and $X_2$?
Let $(X,Y)$ have the joint discrete distribution such that $X \mid Y=y \sim \text{Binomial} \quad (y, 0.5)$ and $Y \sim \text{Poisson}(\lambda), \lambda > 0$, where $\lambda$ is an unknown parameter. Let $T = T(X,Y)$ be any unbiased estimator of $\lambda$. Then
