Finding likelihood function of exponential distribution

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?

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

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 $$\mu$$.

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

• Also $n/{\sum_{i=1}^{n}x_{i}} = 1/\bar x$ Commented Oct 13, 2022 at 14:52
• 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 Commented Oct 13, 2022 at 16:21
• @AlexForester it is a nested differentiation, first you take the derivative of the logarithm and then of the $-\mu$. Hope my answer helps Commented Oct 13, 2022 at 17:09
• 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.
– whuber
Commented Oct 13, 2022 at 18:07