To find the method of moments estimator, we first need to calculate the population mean and variance.
Population mean:
[mathjax]\begin{aligned} E[X_i] &= \int_{-\infty}^{\infty} xf_{a,\lambda}(x) dx \\ &= \int_{a}^{\infty} x \lambda e^{-\lambda(x-a)} dx \\ &= a+\frac{1}{\lambda} \end{aligned}[/mathjax]
Population variance:
[mathjax]\begin{aligned} \text{Var}(X_i) &= E[X_i^2] - (E[X_i])^2 \\ &= \int_{-\infty}^{\infty} x^2 f_{a,\lambda}(x) dx - \left(a+\frac{1}{\lambda}\right)^2 \\ &= \int_{a}^{\infty} x^2 \lambda e^{-\lambda(x-a)} dx - \left(a+\frac{1}{\lambda}\right)^2 \\ &= \frac{1}{\lambda^2} \end{aligned}[/mathjax]
Now we set the sample moments equal to the population moments and solve for [mathjaxinline]\, a \,[/mathjaxinline]:
[mathjax]\begin{aligned} \overline{X}_n &= a+\frac{1}{\lambda} \\ \frac{1}{n} \sum_{i=1}^{n} X_i^2 &= \frac{1}{\lambda^2} \end{aligned}[/mathjax]
Solving for [mathjaxinline]\, a \,[/mathjaxinline] in the first equation, we get:
[mathjax]\begin{aligned} a &= \overline{X}_n - \frac{1}{\lambda} \end{aligned}[/mathjax]
Substituting this into the second equation and solving for [mathjaxinline]\, \lambda \,[/mathjaxinline], we get:
[mathjax]\begin{aligned} \frac{1}{n} \sum_{i=1}^{n} X_i^2 &= \frac{1}{\left(\overline{X}_n - a\right)^2} \\ \lambda &= \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - a\right)^2}} \end{aligned}[/mathjax]
Therefore, the method of moments estimator for [mathjaxinline]\, a \,[/mathjaxinline] is [mathjaxinline]\, \hat{a} = \overline{X}_n - \frac{1}{\hat{\lambda}} \,[/mathjaxinline] and the method of moments estimator for [mathjaxinline]\, \lambda \,[/mathjaxinline] is [mathjaxinline]\, \hat{\lambda} = \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - \hat{a}\right)^2}} \,[/mathjaxinline].
[mathjaxinline]\, X_ i \,[/mathjaxinline] follows a shifted exponential distribution with parameters [mathjaxinline]\, a \in \mathbb {R} \,[/mathjaxinline] and [mathjaxinline]\, \lambda > 0 \,[/mathjaxinline]. That means each [mathjaxinline]\, X_ i \,[/mathjaxinline] has density
[mathjax]f_{a, \lambda }(x) = \lambda e^{-\lambda (x-a)} \mathbf{1}\{ x \geq a\} , \quad x \in \mathbb {R}.[/mathjax]
(If applicable, enter barX_n for [mathjaxinline]\, \overline{X}_ n \,[/mathjaxinline] and bar(X_n^2) for [mathjaxinline]\, \overline{X_ n^2} \,[/mathjaxinline].)
Method of moments estimator [mathjaxinline]\, \hat a = \,[/mathjaxinline]
3 answers
in normal math notation
To find the method of moments estimator for the shifted exponential distribution with parameters [mathjaxinline]\, a \,[/mathjaxinline] and [mathjaxinline]\, \lambda \,[/mathjaxinline], we first need to calculate the population mean and variance. The density function of [mathjaxinline]\, X_i \,[/mathjaxinline] is given by
[mathjax]\, f_{a, \lambda }(x) = \lambda e^{-\lambda (x-a)} \mathbf{1}\{ x \geq a\} , \quad x \in \mathbb {R}.\,[/mathjax]
Population mean:
[mathjax]\, E[X_i] = \int_{-\infty}^{\infty} xf_{a,\lambda}(x) dx = \int_{a}^{\infty} x \lambda e^{-\lambda(x-a)} dx = a+\frac{1}{\lambda}.\,[/mathjax]
Population variance:
[mathjax]\, \text{Var}(X_i) = E[X_i^2] - (E[X_i])^2 = \int_{-\infty}^{\infty} x^2 f_{a,\lambda}(x) dx - \left(a+\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}.\,[/mathjax]
Now we set the sample moments equal to the population moments and solve for [mathjaxinline]\, a \,[/mathjaxinline]:
[mathjax]\, \overline{X}_n = a+\frac{1}{\lambda}, \quad \frac{1}{n} \sum_{i=1}^{n} X_i^2 = \frac{1}{\lambda^2}.\,[/mathjax]
Solving for [mathjaxinline]\, a \,[/mathjaxinline] in the first equation, we get:
[mathjax]\, a = \overline{X}_n - \frac{1}{\lambda}.\,[/mathjax]
Substituting this into the second equation and solving for [mathjaxinline]\, \lambda \,[/mathjaxinline], we get:
[mathjax]\, \lambda = \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - a\right)^2}}.\,[/mathjax]
Therefore, the method of moments estimator for [mathjaxinline]\, a \,[/mathjaxinline] is [mathjaxinline]\, \hat{a} = \overline{X}_n - \frac{1}{\hat{\lambda}} \,[/mathjaxinline] and the method of moments estimator for [mathjaxinline]\, \lambda \,[/mathjaxinline] is [mathjaxinline]\, \hat{\lambda} = \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - \hat{a}\right)^2}} \,[/mathjaxinline].
[mathjax]\, f_{a, \lambda }(x) = \lambda e^{-\lambda (x-a)} \mathbf{1}\{ x \geq a\} , \quad x \in \mathbb {R}.\,[/mathjax]
Population mean:
[mathjax]\, E[X_i] = \int_{-\infty}^{\infty} xf_{a,\lambda}(x) dx = \int_{a}^{\infty} x \lambda e^{-\lambda(x-a)} dx = a+\frac{1}{\lambda}.\,[/mathjax]
Population variance:
[mathjax]\, \text{Var}(X_i) = E[X_i^2] - (E[X_i])^2 = \int_{-\infty}^{\infty} x^2 f_{a,\lambda}(x) dx - \left(a+\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}.\,[/mathjax]
Now we set the sample moments equal to the population moments and solve for [mathjaxinline]\, a \,[/mathjaxinline]:
[mathjax]\, \overline{X}_n = a+\frac{1}{\lambda}, \quad \frac{1}{n} \sum_{i=1}^{n} X_i^2 = \frac{1}{\lambda^2}.\,[/mathjax]
Solving for [mathjaxinline]\, a \,[/mathjaxinline] in the first equation, we get:
[mathjax]\, a = \overline{X}_n - \frac{1}{\lambda}.\,[/mathjax]
Substituting this into the second equation and solving for [mathjaxinline]\, \lambda \,[/mathjaxinline], we get:
[mathjax]\, \lambda = \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - a\right)^2}}.\,[/mathjax]
Therefore, the method of moments estimator for [mathjaxinline]\, a \,[/mathjaxinline] is [mathjaxinline]\, \hat{a} = \overline{X}_n - \frac{1}{\hat{\lambda}} \,[/mathjaxinline] and the method of moments estimator for [mathjaxinline]\, \lambda \,[/mathjaxinline] is [mathjaxinline]\, \hat{\lambda} = \frac{1}{\sqrt{\frac{1}{n} \sum_{i=1}^{n} X_i^2 - \left(\overline{X}_n - \hat{a}\right)^2}} \,[/mathjaxinline].
1 answer