# Probability (cdf question)

1. Dec 17, 2007

### mattmns

[SOLVED] Probability (cdf question)

Here is the problem from the book
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Let X be a random variable with distribution function (cdf)

$$F(x)=\begin{cases} 0 &\text{for } x\geq 0\\ \frac{x}{8} & \text{for } 0 \leq x < 1\\ \frac{1}{4} + \frac{x}{8} & \text{for } 1 \leq x < 2\\ \frac{3}{4} + \frac{x}{12} & \text{for } 2 \leq x < 3\\ 1 & \text{for } x \geq 3\end{cases}$$

Calculate $$P(1 \leq X \leq 2)$$.
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This should be a simple question, but I am getting a different answer than my book is, and I believe the book is wrong (how could I be wrong? :tongue2:).

Here is what I did:

\begin{align*} P(1 \leq X \leq 2) & = P(1 \leq X < 2) \\ & = P(X < 2) - P(X \leq 1) \\ & = \frac{1}{4} + \frac{2}{8} - \left(\frac{1}{4} + \frac{1}{8}\right) \\ & = \frac{1}{8} \end{align*}

We could also get the same answer by finding the pdf and then integrating it over the interval (1,2).

The book I have gives an answer of $$\frac{19}{24}$$.

Here is what they did:

\begin{align*} P(1 \leq X \leq 2) & = P(X\leq 2) - P(X < 1) \\ & = F(2) - \lim_{x\to 1^-}F(x) \\ & = \frac{11}{12} - \frac{1}{8} \\ & = \frac{19}{24}\\ \end{align*}

In my opinion this seems doubly wrong. They used the wrong function on both, but at least they were consistent I suppose. Am I being silly here and missing something, or is the book wrong? Thanks!

edit... I am looking at my book, and they have $$P(a < X \leq b) = F(b) - F(a)$$. Now I won't argue with this, but the way it is used in the above example seems completely counterintuitive to me. I will admit my use of the cdf may be dubious for P(X < 2), but it feels right. Maybe the book is right after all. Thoughts?

Last edited: Dec 17, 2007
2. Dec 17, 2007

### Dick

There are two delta functions in your pdf as well as the continuous part. And they are both included in [1,2]. So P(1<X<2) is not equal to P(1<=X<=2).

3. Dec 17, 2007

### mattmns

So you are saying we have a mixed distribution, and that at least one of P(X = 1), P(X = 2) is non-zero? So from looking at the graph of the cdf, I see that P(X = 1) is 1/4 and P(X = 2) is 5/12.

So if we take what I have, 1/8, and add the points I forgot we get 19/24 which is the book's answer. Thanks, I guess this problem was a little more complicated than I had originally thought.

I guess I should take a closer look at what the cdf actually looks like

4. Dec 17, 2007

### Dick

Yes, mind the discontinuities.