I am curious as to how do find the likelihood function for the exponential distribution with parameters such as this:

$$X \sim \exp(\beta- \mu) $$

With the following assumptions,

  • $\beta$ is known
  • $\mu< \beta$
  • We have an access to an i.i.d. sample $X = X_1, \dots, X_N$ of size $N$

Do we just substitute $(\beta-\mu)$ into the $\lambda$ for the pdf of exponential distribution like below:

$f(x)=\lambda \exp⁡(-\lambda x)$ to become $f(x; \mu)=(\beta- \mu)\exp((\beta-\mu)x) ? $

Then just find the likelihood function? Wouldn't it be complicated to find the MLE of it as well?

  • $\begingroup$ I think yes you plug $b-\mu$ for $\lambda$ and calculate the MLE as usual by paying attention to the restriction $\mu < b$ $\endgroup$
    – Fiodor1234
    Commented Oct 13, 2022 at 14:26

1 Answer 1


If you write as $\lambda = b - \mu$ then you can rewrite the exponential distribution as $f(x;b,\mu_ = (b-\mu)e^{-(b-\mu)x}$.

Now, you have access to iid sample $x_{1}, x_{2},..., x_{n},$ you can write the likelihood function

$$L(\mu|b,x_{1}, x_{2},..., x_{n}) = \prod_{i=1}^{n}(b-\mu)e^{-(b-\mu)x_{i}}$$

then $l(\mu|b,x_{1}, x_{2},..., x_{n}) = log(b-\mu)^{n} - (b-\mu)\sum_{i=1}^{n}x_{i}$, for which we can take the first derivative and equate it to zero so we can maximize it with respect to $\mu$.

$$\frac{dl(\mu|b,x_{1}, x_{2},..., x_{n})}{d\mu} = -\frac{n}{b-\mu}+\sum_{i=1}^{n} x_{i} = 0 \Rightarrow \mu = b - \frac{n}{\sum_{i=1}^{n}x_{i}} = b - \frac{1}{\bar{x}}, \ \ \mu < b, \ \ b > 0 $$

  • 1
    $\begingroup$ Also $n/{\sum_{i=1}^{n}x_{i}} = 1/\bar x$ $\endgroup$
    – Firebug
    Commented Oct 13, 2022 at 14:52
  • $\begingroup$ Sorry if it's a dumb question, but when you differentiate the log likelihood, isn't it supposed to be n/(b−μ)-∑x ? Though it would give the same maximum $\endgroup$ Commented Oct 13, 2022 at 16:21
  • $\begingroup$ @AlexForester it is a nested differentiation, first you take the derivative of the logarithm and then of the $-\mu$. Hope my answer helps $\endgroup$
    – Fiodor1234
    Commented Oct 13, 2022 at 17:09
  • 2
    $\begingroup$ Unfortunately this answer is incorrect as well as confusing. Because "$\beta$ is known," it is evident that $\mu,$ not $\beta,$ is to be estimated. Crucially, you must include an indicator function for the constraint $\mu \lt\beta.$ See stats.stackexchange.com/… for our threads on this topic. $\endgroup$
    – whuber
    Commented Oct 13, 2022 at 18:07

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