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I am quite stumped by the following problem. The usual log-likelihood route with differentiation doesn't work. The problem is as follows:

Let $X_1,...,X_n$ be i.i.d. random variables from a distribution with p.d.f.:

$$f(x; \theta_1, \theta_2) = \frac{1}{\theta_2} \text{exp}(-\frac{(x - \theta_1)}{\theta_2}) \: \text{for } x \geq \theta_1$$

with parameters $\theta_1 \in R$ and $\theta_2 > 0$.

Find maximum likelihood estimators of $\theta_1$ and $\theta_2$

Any hints or insights?

What I tried:

enter image description here

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  • $\begingroup$ If you set the partial derivatives to 0 can't solve the 2 equations in 2 unknowns. If the solution is not unique verify which one is a global maximum. If the solution is unique toy should still be able to verify that it is a maximum. $\endgroup$ Commented Apr 12, 2018 at 3:46
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    $\begingroup$ In the middle of your notebook paper you have $\log{L}=-\frac{\sum _{i=1}^n x(i)}{\theta _2}+\frac{\theta _1 n}{\theta _2}-n \log \left(\theta _2\right)$. Note that this function can increases with increasing $\theta_1$ and the largest value $\theta_1$ can possibly take is the minimum $x_i$. That should tell you what the maximum likelihood estimator of $\theta_1$ is. $\endgroup$
    – JimB
    Commented Apr 12, 2018 at 4:29
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    $\begingroup$ There seems to be an error in your last calculation. Shouldn't the last term be $-n\theta_1/\theta_2^2$ $\endgroup$
    – RDK
    Commented Apr 12, 2018 at 4:39
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    $\begingroup$ You should review your basic material on finding global and local maxima. You should have been taught that setting derivatives to 0 won't always find the maximum (e.g. try it on "find the maximum of $f(x)=1/x$ in the interval $1\leq x\leq 2$"). If you try to set the derivative to $0$ you get nowhere. You've forgotten some important facts! As a first step, consider $\theta_2=1$ and try drawing a picture of the log-likelihood as a function of $\theta_1$ (keeping in mind the relationship between $\theta_1$ and any observation). $\endgroup$
    – Glen_b
    Commented Apr 12, 2018 at 6:20

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To close this one:

This is the pdf of a shifted Exponential random variable, $Y \sim \text{Exp}(\theta_2)$ (scale parameter), and $X = Y + \theta_1$.

From the condition $X\geq \theta_1$, the MLE is immediately obtained as

$$\hat \theta_{MLE} = X_{(1)}$$

i.e. the minimum order statistic and so the actual estimate will be the minimum value available in the sample. This can be verified by directly inspecting the log-likelihood as suggested in a comment.

Then apply "concentrated ML" estimation, i.e. plug the minimum $x$-value in place of $\theta_1$ and optimize as usual with respect to $\theta_2$ only.

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