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Let $(\Omega,\mathscr{F},P)$ be a probability space and $X$ be a random variable that takes values in $\mathbb{N}$. Define $$q(n)\equiv P(X=n)\quad n\in\mathbb{N}$.$$ So $q$ is just the probability mass. $q$ can be viewed as a mapping from $\mathbb{N}$ to $[0,1]$. Then $q(X):\Omega\rightarrow[0,1]$ is another random variable. What is the interpretation of $q(X)$?

Assume we have a sequence of random variables $\{X_k\}$ and $q_k$ is the mass function for $X_k$. Suppose $X_n\rightarrow X$ a.s., is it true that $q_k(X_k)$ converges a.s.?

Thanks

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  1. $q(X)$ is the probability that $X$ takes the value that it happens to take.
  2. Do you want to assume that $X_k$ takes values in $\mathbb N$? If $X_k$ are continuous random variables, you'd have $q_k(X_k) = 0$ a.s.
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  • $\begingroup$ 1. Yes, literally this is what it means. But it is a little hard for me to understand. $\endgroup$
    – user111463
    Apr 22 '15 at 6:54
  • $\begingroup$ 2. Yes, I wan to consider the case that $X_k$ takes discrete values. Otherwise, if $X_k$ is continuous, then I think we should consider $q_k$ as density instead of $P(X_k=n)$. $\endgroup$
    – user111463
    Apr 22 '15 at 6:59

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