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So I need a little help with this please. I'm given N measurements of a signal $Y_{i} = A + v_{i}, i = 1,...,N$, where $v_{i}$ is measurement noise with the exponential pdf $f_{v}(v) = e^{-v}, v \geq 0$. My task is to construct MLE for constant value $A$. The likelihood function should be:

$L_{(Y;A)} = \prod_{i=1}^N e^{-v_{i}}$

and

$l_{(Y;A)} = ln(L_{(Y;A)}) = -\sum_{i=1}^N v_{i} = -\sum_{i=1}^N (Y_{i} - A)$

Now since my job is to estimate parameter A, the next step should (?) be:

$\frac{d}{dA}l(Y;A) = 0 = - \sum_{i=1}^N(-1) = N$

but obviously this doesn't make sense. I'm really new to this and any help would be much appreciated. Thanks

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Not every optimization problem is solved by taking derivatives. And, the PDF is actually $$f_V(v)=e^{-v}\mathbb{I}(v\geq 0)$$

So, we try to maximize $$L=\prod_{i=1}^N e^{A-Y_i}\mathbb{I}(Y_i\geq A)$$

Increasing $A$ monotonically increases the first multiplicand, regardless of $Y_i$. But, there is a limit that we can increase $A$, since the second expression, i.e. the indicator shouldn't be $0$. That means $A\leq \min(Y_i)$. And, the ML estimate will be $\min(Y_i)$ because $A$ being as large as possible is a scenario that we want.

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    $\begingroup$ This actually solves everything, just haven't seen this approach up until now. Thank you. $\endgroup$ – Nikola Petrevski Jan 5 at 17:46
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Hints:

  1. You are implicitly using the fact that the likelihood is zero when any $v_i <0$. You should make this explicit

  2. Your calculations do make sense and suggest that the likelihood is an increasing function of $A$, i.e. $A$ should be as large as possible

The key to this is as possible

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  • $\begingroup$ (1) Right, I forgot about that, I edited my question, thanks. $\endgroup$ – Nikola Petrevski Jan 5 at 17:45
  • $\begingroup$ So the answer is that the likelihood is maximised when $A$ is as large as possible, but $A$ cannot be larger than any of the $Y_i$. So the maximum likelihood estimator is $\hat A = \min Y_i$ $\endgroup$ – Henry Jan 5 at 23:39

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