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This doubt is from the book Stochastic Processes by Ross..Chapter 1, subsection Random variables.

The distribution function F of the random variable X is defined for any real number x by

F(x) = P{X$\leq$x} = P{X $\epsilon$ (-$\infty$,x]}

I have not understood why X $\epsilon$ (-$\infty$,x] and not [0,x]

How is X $\epsilon$ (-$\infty$,x] to be practically interpreted? Request help understand

SAK
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1 Answers1

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Let us consider a particular example. Suppose that a random variable $X$ is equal to $-1$ almost surely. What is the probability that the value of this random variable is less than or equal to some positive number $x$, i.e. $P(X\le x)$, where $x>0$? This probability is not equal to the probability $P(X\in[0,x])$, it is equal to the probability $P(X\in(-\infty,x])$.

I hope this helps.

Cm7F7Bb
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  • Yes Thank you,,,, I was taking x as to be events and not general random numbers... so could not figure out how events could be negative. – SAK Oct 31 '16 at 14:46
  • @SAK You're welcome! Yes, $X$ is a random variable and $F(x)$ denotes the probability of the event that the random variable takes value in the set $(-\infty,x]$, i.e. the probability of the event ${\omega\in\Omega:X(\omega)\in(-\infty,x]}$. – Cm7F7Bb Oct 31 '16 at 14:48