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The lifetime of computer monitors has a exponential distribution where the expected value can be written as: $\mu(s) = \frac{\beta}{s}$

Where $s$ is how bright the monitor is and both $s$ and $\beta$ are >0

How do I estimate the maximum-likelihood of parameter $\beta$?

I know that the log-likelihood for a exponential distribution is:

$l(\lambda) = n(\ln(\lambda) - \lambda\bar{t})$

And taking the derivative and equals it to zero yields:

$\frac{dl(\lambda)}{d\lambda} = n(\frac{1}{\lambda} - \bar{t}) = 0$

$\bar{t} = \frac{1}{\lambda}$

Here is where I get stuck.

EDIT 1

I have data for both the lifetime of the computer monitors and their brightness.

utobi
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0xcc
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3 Answers3

9

To estimate the $\beta$ by the maximum likelihood method, let $Y_1,\ldots, Y_n$ be the sample of lifetimes, with $Y_i\sim \text{Exp}(\beta/s_i)$, s.t. $E(Y_i) = \beta/s_i$ and independently for each $i$, where $s_i$ is the monitor brightness.

Then the likelihood function for $\beta$ is given by

$$ L(\beta) = \prod_{i=1}^n (s_i/\beta) e^{-\frac{1}{\beta}s_i Y_i} \propto \beta^{-n} e^{-\frac{1}{\beta}\sum_is_iY_i}. $$

I leave the rest to you...

utobi
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    Thank you, but there is something i don't understand, first, you plug in that $\lambda = \frac{\beta}{s}$, should it not be the inverse? That is $\lambda = \frac{s}{\beta}$ since the mean of a exponential distribution is: $\frac{1}{\lambda}$

    Also if i continue with your calculations i get that the log-likelihood is : $l(\beta) = n \ln(\beta) \sum_i{\frac{Y_i}{s_i}}$

    Taking it's derivative and setting it equal it to zero yields: $l'(\beta) = \frac{1}{\beta} n\sum_i{\frac{Y_i}{s_i}} = 0 $, which seems wrong?

     – 0xcc Nov 24 '22 at 18:23
  • You are right @0xcc, I was using a different reparametrization, not the one you are working with. See the updated answer. – utobi Nov 24 '22 at 18:43
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    Ok so i think i got it, the expression for the log-likelihood is: $-\ln(\beta) - \frac{\sum_i{Y_i s_i}}{\beta}$ So i want to try maximize that with respect to $\beta$ given my data? – 0xcc Nov 24 '22 at 19:46
  • almost...its $-n\ln\beta-\cdots$ – utobi Nov 24 '22 at 19:47
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    Wow now i got it, @utobi you are fantastic! Thank you. – 0xcc Nov 24 '22 at 19:48
  • Glad to see it helped! – utobi Nov 24 '22 at 19:54
4

$ L(\beta) = P(\mathrm{Data} \mid \beta) = \prod_i f(x_i) $ and $\lambda = \frac{1}{\mu} = \frac{s}{\beta}$

Log is increasing so maximizing log is same as maximizing likelihood: $\ell(\beta) = \sum \log f(x_i).$

$ \log f(x_i) = \log \lambda e^{-\lambda x_i} = \log s - \log \beta - \frac{s}{\beta}x_i$

and

\frac{d}{d \beta} \log f = -\frac{1}{\beta} - \frac{s}{\beta^2}x_i.

So,

\frac{d}{d \beta}\ell(\beta) = -\frac{n}{\beta} - \frac{s}{\beta^2}\sum x_i = 0

gives

\hat{\beta} = \frac{s}{n}\sum x_i = s \bar{x}.

Note: the second derivative of $\ell$ is positive since $x_i, s, \beta>0$ so LL is concave.

Hunaphu
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3

Assise from using a direct derivation, leading to a closed form expression, you can do this with a generalized linear model a description is given in the question

Fitting exponential (regression) model by MLE?

With that approach you have more flexibility (e.g. change the function $\mu(s)$)

Example

In R you would use the function glm with the family gamma and specify a dispersion parameter dispersion=1 (the exponential distribution is a special case of the gamma distribution for which the dispersion parameter is equal to 1).

### generate and plot data

set.seed(1)
n = 100
x = runif(n,0,10)
mu = 2/x
rate = 1/mu
y = rexp(n,rate)
plot(x,y)


fit glm model with Gamma distribution that has k=1 (ie. the exponential distribution)


fit = glm(y ~ 0+x, family = Gamma())
summary(fit,dispersion=1)


This gives as result


a rate parameter of lambda = 0.49868 x


which relates to


a mean that is mu = 1/lambda = 2.005294/x
Sextus Empiricus
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