See this link at Page-$07$.

If the stochastic process is discrete-valued, then a collection of probability mass functions can be used to specify the stochastic process:

$$P_{X_1···X_k}(x_1, x_2, · · · , x_k) = P(X_1 = x_1, X_2 = x_2, ··· , X_k = x_k)$$

How does the above r.p. notation read in plain English?

As far as I can tell:

Probability of the Random Process $X(k, \zeta)$ is equal to the joint probability of the random variables $X_1, X_2, ... , X_k$.

Note. $\zeta$ is the sample space.

But, why is it written as $P(X_1 = x_1, X_2 = x_2, ··· , X_k = x_k)$?

It seems that $X_1$ can't take the value $x_2$? Why?

Aren't $x_1$ and $x_2$ both members of sample space $\zeta$ ?


The lower-case $x_1, x_2, \ldots, x_k$ are dummy variables. They represent generic values for the corresponding random variables $X_1, X_2,\ldots, X_k$, the same way that $x$ and $y$ are generic variables in the joint density $f(x,y)$ of a bivariate pair $(X,Y)$. The dummy variables have different subscripts so we can keep straight which dummy variable is associated with which random variable.

In equations involving $x_1,x_2,\ldots,x_k$, unless explicitly restricted, these variables are understood to take any legal value: For example, in the equation (which is actually defining the LHS) $$ P_{X_1···X_k}(x_1, x_2, · · · , x_k) = P(X_1 = x_1, X_2 = x_2, ··· , X_k = x_k) $$ you should read the implicit "for all possible values of $x_1,x_2,\ldots,x_k$".

  • $\begingroup$ Are rest of my conceptions correct? $\endgroup$ – user366312 Oct 11 '18 at 7:08
  • $\begingroup$ @yahoo.com Yes, the probability mass function $P_{X_1···X_k}(x_1, x_2, · · · , x_k)$ is shorthand for the joint probability $P(X_1 = x_1, X_2 = x_2, ··· , X_k = x_k)$. The assertion is that to define the random process, it is enough to specify the collection of p.m.f.'s as $k$ ranges over all positive integers. $\endgroup$ – grand_chat Oct 11 '18 at 7:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.