Finding likelihood function of exponential distribution

Question

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?

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 $\mu$.

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