# Statistical estimation: how is the following probability distribution defined?

I am currently reading Chapter 7 of Stephen Boyd's textbook. From Stephen Boyd's optimization textbook, it says From my understanding,

• If $$X$$ is a random variable on a probability space $$(\Omega, \mathcal{A}, \mu)$$, then $$X$$ induces a probability measure on $$\mathbb{R}$$ called the probability distribution, $$\mu(A) = \Pr(X \in A)$$, $$A \in \mathcal{A}$$.

• A distribution function of $$X$$ is the function $$F(x) = \Pr(X\leq x)$$, also known as the CDF.

• The density of $$F$$ is the function $$f$$ such that $$F(x) = \int\limits_{-\infty}^x f(y) dy$$

So what does probability distribution "indexed" by a vector mean in this context? Also I am troubled by the notation $$p_x(\cdot)$$. In my experience, almost all densities are written as $$f_X(x)$$, where $$X$$ is the random variable. But here $$x$$ is just a vector. In the entire chapter, the notion of a random variable is never even brought up, yet curiously a probability distribution etc is still able to be defined.

Can someone help me understand the meaning of the very first sentence and offer an example of what a concrete probability distribution and density $$p_x(\cdot)$$ in this context?

Consider the following example. Let $$Y\sim\exp(\lambda)$$, where $$\lambda\in \mathbb{R}_{>0}$$. The value of parameter $$\lambda$$ is unknown. So one considers the family of probability distributions on $$\mathbb{R}$$ indexed by $$\lambda$$ s.t. $$p_{\lambda}(y)=\lambda\exp(-\lambda y)$$, $$y\ge 0$$. Letting $$P_{\lambda}$$ denote the distribution corresponding to $$p_{\lambda}$$, i.e. $$P_{\lambda}(A)=\int_A p_{\lambda}(y)\,dy$$, this family can be written as $$\{P_{\lambda}:\lambda\in(0,\infty)\}$$. The log-likelihood in this case is given by $$l(\lambda)=\ln p_{\lambda}(y)=\ln\lambda-\lambda y.$$
If we are given $$n$$ i.i.d. copies of $$Y$$, i.e. $$(Y_1,\ldots,Y_n)$$, then $$p_{\lambda}(y)=\prod_{i=1}^n \lambda \exp(-\lambda y_i)$$, where $$y\equiv(y_1,\ldots,y_n)\ge 0$$. The log-likelihood in this case becomes $$l(\lambda)=n\ln\lambda-\lambda\sum_{i=1}^n y_i.$$