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https://www.statlect.com/asymptotic-theory/convergence-in-probability

Let X be a discrete random variable with support $R_x = \{0,1\}$ and probability mass function

$p_X(x) = \begin{cases} \frac{1}{3}, \; if \; x = 1,\\ \frac{2}{3}, \; if \; x = 0,\\ 0, \; otherwise, \end{cases}$

Consider a sequence of random variables $\{X_n\}$ whose generic term is $\;\;X_n = (1+\frac{1}{n})X\;\;$. To prove that $\{X_n\}$ converges in probability to $X $ they write $|X_n-X| = |(1+\frac{1}{n})X- X|=|\frac{1}{n}X|$. My question is how can they cancel random variable?

e.g. If I take $X_1 = X$ and $X_2 = X$ then $X_1 -X_2 \ne 0$ right?

It may be very trivial but do explain when random variables can be cancelled.

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    $\begingroup$ Random variables are functions $\Omega\to\mathbb R$ and an expression like $X_1=X$ must be red as: $X_1(\omega)=X(\omega)$ for each $\omega\in\Omega$. If also $X_2(\omega)=X(\omega)$ for every $\omega\in\Omega$ then $(X_1-X_2)(\omega):=X_1(\omega)-X_2(\omega)=0$ for every $\omega\in\Omega$. Then you can write $X_1-X_2=0$ but this in the understanding that RHS $0$ denotes the zero-function on $\Omega$. This $0$ is a degenerated random variable. $\endgroup$ – drhab May 6 '17 at 10:13
  • $\begingroup$ @drhab I got your point. Thanks. I have updated the problem. Can you please also clarify this point? $\endgroup$ – hunch May 6 '17 at 16:59

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