Given CDF find proabililty

• Jun 15th 2011, 02:13 PM
jwxie
Given CDF find proabililty
Hi guys, I feel very uncomfortable with PDF, CDF, PMF.

Problem:

Quote:

The distribution function of the random variable X is given:

$\displaystyle F(x) = \begin{cases}0 & \text{ if } x < 0\\x/2 & \text{ if } 0\leq x < 1\\2/3 & \text{ if } 1\leq x<2\\11/12 & \text{ if } 2\leq x < 3 \\1 & \text{ if } 3\leq x\end{cases}$

a) Plot the distribution function.
b) What is P { X > 1/2} ?
c) What is P{2 < X <= 4} ?
d) What is P {X <3 }?
e) What is P{X = 1}?
I think the plot looks like this:
http://dl.dropbox.com/u/14655573/lat...28.resized.JPG

In this problem we mix both discrete and continuous random variable (R.V.).

For the discrete case, we have PMF and CMF, which they both can give the probability. For continuous, we can only find probability via CMF because P {X = a} (of any particular point) is zero..

We know that $\displaystyle F(x) = P(X \leq x) = \int_{-\infty }^{x} f(t)dt$

and for discrete we have $\displaystyle F(x) = \sum_{x_{i}\leq x} p(x_{i})=\int_{-\infty }^{x}f(t)dt$

For $\displaystyle P(a\leq X \leq b) = \int_{a}^{b}f(x)dx=F(b)-F(a)$

=============

For (b) What is P { X > 1/2}:

I think the solution is simply 1 - P {x <=1/2} because the probability has to add up to 1.

For (c) P{2 < X <= 4}
Do we just take F(4) - F(2)? Since this is a continuous function, P (x = a) is zero, so is it true that P {2 <= X <= 4} = P {2 < X < 4} = P {2 <= X < 4} ???

For d) What is P {X <3 } ...... similarly, this problem does not include the = sign. Is it equal to P {X <=3 }?? I heard my instructor said something about " P {X < 3- }....
I am not sure how to answer.

For (e) P {X = 1}.... the limit is 2/3 if 1 <= x < 2, so this is a discrete case, right? I'd say P = 2/3, right?

==== I always have difficult time to ask the right questions......but for now I am going to stop here.... thank you for reading and I appreciate for the help! Thanks.
• Jun 15th 2011, 03:32 PM
Plato
Re: Given CDF find proabililty
This question is very hard to answer because it is a notation nightmare.

Let $\displaystyle F(a - ) = \lim _{x \to a^ - } F(x)$, that is the left-hand limit at $\displaystyle x=a$.

Likewise, $\displaystyle F(a + ) = \lim _{x \to a^ + } F(x)$, that is the right-hand limit at $\displaystyle x=a$.

Now I can give you nine rules:
$\displaystyle \begin{array}{rrcl} {1.} & {P(X \leqslant a)} & = & {F(a)} \\ {2.} & {P(X < a)} & = & {F(a - )} \\ {3.} & {P(a < X \leqslant b)} & = & {F(b) - F(a)} \\ {4.} & {P(X = a)} & = & {F(a) - F(a - )} \\ {5.} & {P(a < X < b)} & = & {F(b - ) - F(a)} \\{6.} & {P(a \leqslant X \leqslant b)} & = & {F(b) - F(a - )} \\{7.} & {P(a \leqslant X < b)} & = & {F(b - ) - F(a - )} \\{8.} & {P(a < X)} & = & {1 - F(a)} \\{9.} & {P(a \leqslant X)} & = & {1 - F(a - )} \\ \end{array}$

So to answer (b) $\displaystyle 1-F(1/2)$.
• Jun 15th 2011, 03:47 PM
jwxie
Re: Given CDF find proabililty
Hi Plato, thanks for the help. I agree.
For #3, a is not included, so why is it F(a) ?
and F(a) without the + sign means both limits right?
• Jun 15th 2011, 04:06 PM
Plato
Re: Given CDF find proabililty
Quote:

Originally Posted by jwxie
For #3, a is not included, so why is it F(a) ?
and F(a) without the + sign means both limits right?

I almost did not post a reply to this query for this reason.
I have written at least twenty pages on this topic.
Do you agree that the complement of $\displaystyle ( - \infty ,a]$ is $\displaystyle (a,\infty )~?$
If you do understand that the answer is clear.

IF NOT, I must remind you that that this is it a tutorial service.