Solving a Maximum Likelihood Estimation with an exponential distribution I need someone's insight on applying a MLE for an exponential distribution. In a finance paper, I have the following:
$\displaystyle d_i \sim \frac{\epsilon_i}{\lambda_i}$ where $\epsilon_i$ is i.i.d. exponentially distributed with parameter $= 1$.
and $i=1,\ldots,n$. 
$d_i$ are duration time values like time between two events. The $\epsilon$ are not observed. $\lambda_i$ are not observed and must be replaced with estimates from an optimal filter under a $2^k$ states where $k$ can take value $2 \ldots 10$.
Conditional on $\lambda_i$ the $d_i$ have an exponential distribution of $\lambda_i$ with density $p(d_i|\lambda_i) = \lambda_i \exp[-\lambda_i d_i]$ 
The $\epsilon_i$ in $\displaystyle d_i \sim \frac{\epsilon_i}{\lambda_i}$ confuses me in the MLE application. First, is the $\epsilon_i$ relevant in the MLE computation? If yes, how does it influence the likelihood fucntion below:
$$
\mathcal{L}(\lambda,d_1,\dots,d_n)=\prod_{i=1}^n f(d_i,\lambda)=\prod_{i=1}^n \lambda e^{-\lambda d}=\lambda^ne^{-\lambda\sum_{i=1}^nd_i}
$$
 A: You seem to be mixing up two different ideas.  The $n$ random variables are independent but not identically distributed.  If your observation is
$(d_1, d_2, \ldots, d_n)$, then the likelihood function is the product
$$\prod_{i=1}^n \lambda_i \exp(-\lambda_i d_i)$$
which is a function of $n$ unknown parameters 
$\lambda_1, \lambda_2, \ldots, \lambda_n$.  This likelihood function 
has maximum value at
$(\lambda_1, \lambda_2, \ldots, \lambda_n) = (d_1^{-1}, d_2^{-1}, \ldots, d_n^{-1})$.
More likely, your model is that of $n$ independent samples of an 
exponential random variable with unknown parameter $\lambda$. If your
observation is $(d_1, d_2, \ldots, d_n)$, then the
likelihood function is the product
$$\prod_{i=1}^n \lambda \exp(-\lambda d_i)
= \lambda^n\exp\left (-\lambda \sum_{i=1}^n d_i \right)$$
which is a function of the single unknown parameter $\lambda$. The
maximum value of this likelihood occurs at 
$\displaystyle \lambda = \frac{n}{\sum_{i=1}^n d_i } = \frac{1}{\bar{d}}$
where $\bar{d}$ is the sample mean 
$\displaystyle \frac{1}{n}\sum_{i=1}^n d_i$.
