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Let X1, · · · , Xn be an i.i.d. sample from an exponential distribution with the density function. Find the maximum likelihood of $$f(x|β) = \frac{1}{β}\,e^{-x/β}$$

First, I need to find: $$p(x|β) = \prod_{i=1}^n \frac{1}{β} \,e^{-x_i/β}$$ but I'm not sure how I can simplify the equation above.

Here's my attempt: $$p(x|β) = \frac{1}{β^n} \,e^\frac{-\sum_{i=1}^nx_i}{β}$$

Then, I have to find the negative log-likelihood: $$L(x|β) = \ nlog(β) + \frac{\sum_{i=1}^nx_i}{β}$$

Then, I have to differentiate with respect to β to get the MLE. So $$β = \frac{\sum_{i=1}^nx_i}{n}$$

Is this correct? I'm unsure about my simplification.

llamaro25
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    Likelihood based on the sample $x_1,\ldots,x_n$ is $L(\beta\mid x_1,\ldots,x_n)=\prod_{i=1}^n f(x_i\mid \beta)$, a function of $\beta$. See https://math.stackexchange.com/questions/101481/calculating-maximum-likelihood-estimation-of-the-exponential-distribution-and-pr. – StubbornAtom Oct 30 '19 at 17:43
  • When you start simplifying the expression, be careful with the product of the exponentials. It should be $$=\frac{1}{\prod_i \beta_i}e^{\sum_i \frac{-x}{\beta_i}}$$ – Chaos Oct 30 '19 at 17:43
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    It looks fine, notice that the maximum likelihood estimator coincided with the sample mean. Anyway notice that the likelihood function is a function of $\beta$ where the sample $\pmb{X}$ is given as @StubbornAtom mentioned above. – Chaos Oct 30 '19 at 17:59

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