# Maximum Likelihood Estimator with exponential noise

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

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.

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

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

• (1) Right, I forgot about that, I edited my question, thanks. – Nikola Petrevski Jan 5 at 17:45
• 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 $\hat A = \min Y_i$ – Henry Jan 5 at 23:39