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The probability density function of the exponential distribution is defined as

$$ f(x;\lambda)=\begin{cases} \lambda e^{-\lambda x} &\text{if } x \geq 0 \\ 0 & \text{if } x<0 \end{cases} $$

We know that mle of the parameter $\lambda$ is $$\lambda = \frac{n}{\sum\limits_{i=1}^n x_i}$$ According to this solution, it seems $x \geq 0$ has no effect in the solution. My question is that why don't we use the constaint $x \geq 0$ in here? What happens when any of $x_i$'s less than zero?

eet
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  • none of the $X_i$ can be less than zero... – tommik Jun 28 '21 at 13:13
  • ... so $\sum\limits_{i=1}^n x_i$ cannot be less than $0$ and so $\hat \lambda$ will never be negative – Henry Jun 28 '21 at 13:17
  • @Henry, I would say the solution is max[$0$,$\frac{n}{\sum\limits_{i=1}^n x_i}$] but this approach allows some of $x_i$'s zeros. On the other hand, if any of $x_i$'s zero, then the likelihood will be directly zero. – eet Jun 28 '21 at 13:23
  • When $x_i=0$, you have $\lambda e^{-\lambda x}=\lambda$ not $0$. But according to your density $X_i<0$ with probability $0$ so really does not need to be covered – Henry Jun 28 '21 at 13:29
  • @Henry, sorry I should have said less than zero in my previous comment. So, is max[$0$,$\frac{n}{\sum\limits_{i=1}^n x_i}$] true? – eet Jun 28 '21 at 13:45
  • There is zero probability $\sum\limits_{i=1}^n x_i <0$, so you do not need to worry about it – Henry Jun 28 '21 at 13:51

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