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Problem 41

Find the maximum likelihood estimator based on a sample of size \(n\) from the
two-sided exponential distribution with PDF
$$
f(x ; \theta)=\frac{1}{2} e^{-|x-\theta|}, \quad-\infty

Expert verified

The maximum likelihood estimator for \(\theta\) in the given two-sided exponential distribution is the sample median. Furthermore, this estimator is unbiased.

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Chapter 4

Chapter 4

Let \(\bar{X}\) be the mean of a random sample of size \(n\) from \(N(\mu, 25)\). Find the smallest sample size \(n\) such that \((\bar{X}-1, \bar{X}+1)\) is a \(0.95\) level confidence interval for \(\mu .\)

Chapter 4

Let \(X\) and \(Y\) be two independent \(N(0,1)\) random variables. Show that \(X+Y\) and \(X-Y\) are independent.

Chapter 4

Consider a queueing system in which the arrival of customers follows Poisson
process. Let \(X\) be the distribution of service time, which has gamma
distribution;
i.e., the PDF of \(X\) is given by
$$
f(x ; \lambda, r)=\left\\{\begin{array}{ll}
\frac{\lambda(\lambda x)^{r-1} e^{-\lambda x}}{\Gamma(r)}, & 0

Chapter 4

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from uniform distribution on an interval \((0, \theta)\). Show that $\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\( is consistent estimator of \)\theta e^{-1}$.

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