We have $$f(t | \theta)= \frac{1}{\theta}e^{-\frac{t}{\theta}}$$
And we're asked to find out the likelihood function for $\theta$, $t \geq 0$
My thoughts are:
We use the usual process of $\prod_{i=1}^{n}f(t_i|\theta) $ but sometimes you have to add an indicator variable like $I(t_i > \theta)$ of some sort. Is this a case where we do so? And if so what would it be?
Thank you
On
You ought to say at the outset that $$ f(t\mid\theta) = \frac1\theta e^{-t/\theta} \text{ for } t>0. \tag 1 $$
Then you have $$ L(\theta) = \frac1 {\theta^n} e^{-(t_1+\cdots+t_n)/\theta} \text{ for } \theta\ge0. \tag 2 $$ Line $(1)$ above is valid only for $t_1,\ldots,t_n>0.$
Note that I wrote $\text{“}{>}\text{”}$ on line $(1)$ but $\text{“}{\ge}\text{”}$ on line $(2),$ because probability density functions are not defined pointwise, but likelihood functions are.
You can use an indicator to indicate the support of the random variable. This means that your pdf does not actually look like that per se, but it should be $$f(t|\theta) = \frac{1}{\theta}e^{-t/\theta} 1_{[0, \infty)}(t).$$ You can fill this into the usual way of calculating the likelihood.
Note that in this case, this is not so interesting, because the support does not depend on $\theta$. If the support does depend on $\theta$, this can become important.