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Suppose a measurement process is applied to something whose actual value $\mu$ is given by the probability density function $f(x)= \frac{1}{\mu} e^{-\frac{x}{\mu}} $

Suppose we have observations $x_1, x_2 \ldots, x_n$ were taken with precision $\Delta$. This means that each measurement can have a value of $x \pm \Delta$.

Write the likelihood function to find those observations. Find the value that maximizes the likelihood function

I do

$L(\mu)= \frac{1}{\mu} e^{-\frac{x_1}{\mu}} \cdots \frac{1}{\mu} e^{-\frac{x_n}{\mu}}=\frac{1}{\mu^n}e^{\frac{-x_1-x_2- \ldots -x_n}{\mu}}=\frac{1}{\mu^n}e^{\frac{-n\bar{x}}{\mu}}$ but i dont know if im right or i have to consider the $ \pm \Delta$

and for the second i do $\frac{\partial ln(L(\mu))}{\partial \mu}=\frac{n(\bar{x}-\mu)}{\mu^2}=0$ then $\mu=\bar{x}$

HUmbertto Ramso
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  • If your "each measurement can have a value of $x\pm \Delta$" is an absolute statement (rather than say an indication of standard deviation or a confidence interval) then your estimate must be in the interval $[\max x_i - \Delta, \min x_i + \Delta]$ since the likelihood would be zero outside this interval. But then your density would not be $\frac{1}{\mu} e^{-\frac{x}{\mu}}$ but instead a truncated form of this – Henry Nov 27 '23 at 10:32

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