# Thread: Distribution function of a random variable X

1. ## Distribution function of a random variable X

The distribution function of $X$ is given by

$F(b) = \\$

$0$ if $b < 0 \\*$

$\frac{b}{4}$ if $0 \leq b < 1\\$

$\frac{1}{2} +\frac{b-1}{4}$ if $1 \leq b < 2\\$

$\frac{11}{12}$ if $2 \leq b < 3\\$

$1$ if $3 \leq b\\$

I have to find
a) $P({X= i}), i = 1,2,3$
b) $P({\frac{1}{2} < X < \frac{3}{2}})$

For a) I figured possible values for the random variable $X$ were $0,1,2,3$

Doing some subtraction I got:

$P(X=0) = \frac{b}{4}$

$P(X=1) = \frac{1}{4}$

$P(X=2) = \frac{11}{12} - (\frac{1}{2} + \frac{b+1}{4})$

$P(X=3) = 1 - \frac{11}{12} = \frac{1}{12}$

I am confused. What is b? Is it a possible value that the random variable can take on? Also, is part a) the same thing as saying "find the probability mass function"?

And for part b), I am not sure where to begin with this one. Any help is appreciated.

Thank you very much.

-Jame

(Also I apologize if the table is hard to read, that's the nicest I could get it with tex)

2. None of those should contain a variable.
$P(X=1)=\frac{1}{4}$.

It is $P(X=k)=F(k)-F(k-)$ where $\displaystyle F(k-)=
\lim _{b \to k^ - } F(b)$

In other word, find the amount of the 'jump' at k.

3. $F(b) \equiv P(X \le b)$. b is just a dummy variable. Your answers should not have any b's in them.

Also, no, X can assume values other than 0, 1, 2, 3. It can actually attain any value in $(0, 2] \cup \{3\}$.

4. What do you mean by $k-$? I've never seen that notation before.

5. Originally Posted by Jame
What do you mean by $k-$? I've never seen that notation before.
If you read reply #2 very carefully it was indeed explained there.
It is very common notation. F(k-) denotes the left hand limit at k, $\displaystyle\lim _{b \to k^ - } F(b)$.

The cumulative distribution function is continuous from the right.
So we have $F(k)=F(k+)$ where $F(k+)=\displaystyle\lim _{b \to k^ + } F(b)$.

This is convenient notation for CDF's.
$P(X\le k)=F(k)~,$ $P(X $P(j\le X \le k)=F(k)-F(j-)$.

6. Oh yes, I see what you mean. We are concerned with the probability at the "jump". Sorry I was kinda thrown off by the $b$ at first.

The correct probabilities are

$P(X=1) = \frac{1}{4}$
$P(X=2) = \frac{1}{6}$
$P(X=3) = \frac{1}{12}$

Thanks again for the clarification, I see how this works now.

Cheers