- #1
mattmns
- 1,129
- 5
[SOLVED] Probability (cdf question)
Here is the problem from the book
-----------
Let X be a random variable with distribution function (cdf)
<br /> F(x)=\begin{cases}<br /> 0 &\text{for } x\geq 0\\<br /> \frac{x}{8} & \text{for } 0 \leq x < 1\\<br /> \frac{1}{4} + \frac{x}{8} & \text{for } 1 \leq x < 2\\<br /> \frac{3}{4} + \frac{x}{12} & \text{for } 2 \leq x < 3\\<br /> 1 & \text{for } x \geq 3\end{cases}<br />
Calculate P(1 \leq X \leq 2).
----------------
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?
).Here is what I did:
<br /> \begin{align*}<br /> P(1 \leq X \leq 2) & = P(1 \leq X < 2) \\<br /> & = P(X < 2) - P(X \leq 1) \\<br /> & = \frac{1}{4} + \frac{2}{8} - \left(\frac{1}{4} + \frac{1}{8}\right) \\<br /> & = \frac{1}{8}<br /> \end{align*}<br />
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:
<br /> \begin{align*}<br /> P(1 \leq X \leq 2) & = P(X\leq 2) - P(X < 1) \\<br /> & = F(2) - \lim_{x\to 1^-}F(x) \\<br /> & = \frac{11}{12} - \frac{1}{8} \\<br /> & = \frac{19}{24}\\<br /> \end{align*}<br />
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?
Here is the problem from the book
-----------
Let X be a random variable with distribution function (cdf)
<br /> F(x)=\begin{cases}<br /> 0 &\text{for } x\geq 0\\<br /> \frac{x}{8} & \text{for } 0 \leq x < 1\\<br /> \frac{1}{4} + \frac{x}{8} & \text{for } 1 \leq x < 2\\<br /> \frac{3}{4} + \frac{x}{12} & \text{for } 2 \leq x < 3\\<br /> 1 & \text{for } x \geq 3\end{cases}<br />
Calculate P(1 \leq X \leq 2).
----------------
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?
).Here is what I did:<br /> \begin{align*}<br /> P(1 \leq X \leq 2) & = P(1 \leq X < 2) \\<br /> & = P(X < 2) - P(X \leq 1) \\<br /> & = \frac{1}{4} + \frac{2}{8} - \left(\frac{1}{4} + \frac{1}{8}\right) \\<br /> & = \frac{1}{8}<br /> \end{align*}<br />
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:
<br /> \begin{align*}<br /> P(1 \leq X \leq 2) & = P(X\leq 2) - P(X < 1) \\<br /> & = F(2) - \lim_{x\to 1^-}F(x) \\<br /> & = \frac{11}{12} - \frac{1}{8} \\<br /> & = \frac{19}{24}\\<br /> \end{align*}<br />
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:
