# Defining the probability distribution function

Suppose $X$ is a random variable that takes values in $[0,1].$ Is it that the probability distribution of $X$ is defined on $[0,1]$ or whole $R$?

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The probability distribution is defined on the whole of $\mathbb{R}$. Also, you need to understand that, this is given by an integral or sum depending on what $X$ is, whose limits are usually taken to be from $-\infty$ to $x \in \mathbb R$.

In general, any function $F$ is a distribution function if it satisfies the following properties:

• $0 \leq F(x) \leq 1$ for all $x \in \mathbb{R}$

• $F$ is a non-decreasing function of $x$.

• $F(-\infty) := \lim_{x \to -\infty}{F(x)}=0$ and $F(\infty) := \lim_{x \to \infty}{F(x)}=1$

• $F$ is right continuous, i.e. $F(x+)=F(x)$ for all $x \in \mathbb R$.

And, it can be shown that, given $F$ with above properties, there exists a probability space $(\Omega, \mathscr A, P)$ on which a random variable $X$ exists such that $X$ has $F$ as its distribution.

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I think your requires for $F$ may be slightly circular. There are distributions functions which are not defined on $\mathbb{R}$, such as those are defined on $\mathbb{R}^2$ and I see no reason why one cannot be defined on $[0.1]$ –  Henry Jan 13 '12 at 18:19
All I have said is for a one-dimensional random variable, $F$ is necessarily (by convention, if you like), defined on $\mathbb{R}$ –  user21436 Jan 13 '12 at 18:43

There is not a lot of difference.

With a probability space $(\Omega, \mathscr A, P)$, the sample space $\Omega$ must contain all possible events; whether it includes impossible ones too does not matter much so long as you are clear.

If the distribution is defined on the whole of $\mathbb{R}$ but values cannot be less than 0 or more than 1, then you can ignore both of these possibilities. If you keep them then $\Pr(X\lt 0)=0$ and $\Pr(X\gt 1)=0$.

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@Kannappan Sampath has given you all the properties of the distribution function but not really explained what the distribution function is or why it is defined for all $x$. The value of the distribution function $F(x)$ is the probability that the random variable $X$ is no larger than $x$:

For each $x, -\infty < x < \infty$, $F(x) = P\{X \leq x\}$.

For every random variable $X$, $F(x)$ is defined for all real numbers $x$. Your particular $X$ takes on values between $0$ and $1$ only, but we can still talk of $P\{X \leq 5\}$ and say that its value is $1$.
The general property that the limit of $F(x)$ is $1$ as $x \to \infty$ is perfectly true in this case, but it is also true that $F(x) = 1$ for all $x \geq 1$, that is, the limiting value is not approached asymptotically as $x \to \infty$ but is achieved (and held) from $x=1$ onwards as we let $x \to \infty$. Similarly for $\lim_{x \to -\infty} F(x) = 0$ which limiting value is not approached asymptotically.

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This really answers the question! –  user21436 Jan 13 '12 at 21:19