# Why is this the PDF of this CDF?

I have a cumulative distribution function $\displaystyle{% F(x) = \left\lbrace% \begin{array}{ll} 0\,, & \mbox{for}\ x<1 \\[1mm] {1 \over 2}\,\left(x^{2} - 2x + 2\right)\,, & \mbox{for}\ 1 \leq x < 2 \\[1mm] 1\,, & \mbox{for}\ x \geq 2 \end{array}\right.}$

and I'm asked to find the variance for $X$. So when I took the derivative of the CDF I found that $f(x)=\begin{cases}\frac12 \quad \text{for$x=1$}\\ x-1\quad \text{for$1\lt x\lt 2$}\end{cases}$. I understand where the $x-1$ and it's bounds come from, but how does one go about solving for that "jump" at the lower bound where $f(1)=\frac12$. After playing around with the solutions I finally figured it out but I'm not sure how I would have realized that right away. Would I simply just integrate $x-1$ over it's bounds to see if it equals one? I've seen this kind of problem before and I'm trying to avoid making the mistake of leaving the "jump" out. Should I always just integrate over the bounds of a pdf to make sure it equals one everytime? Or is there a way to tell there is a "jump"?

I don't know where you're getting that $1/2$ for $x=1$. The derivative is $0$ for $x < 1$ and for $x > 2$, and $x-1$ for $1 < x < 2$. Since $(x^2 - 2 x + 2)/2$ is $1/2$ at $x = 1$, your CDF is discontinuous there. The conclusion should be that this is not a continuous distribution and therefore does not have a PDF.
• That's what I thought, I'm studying for SOA/CAS exam P and this was one of the practice problems I had and the solution states that this is the pdf. But when I integrate $x-1$ over it's bounds it does only equal $\frac12$. So I would be missing half of the distribution somewhere. I just don't know why at $1$ is where the rest of it is. – TheHopefulActuary Nov 3 '13 at 23:13
• @TheHopefulActuary Your description of the density as $f(x)=\begin{cases}\frac12 \quad \text{for$x=1$}\\ x-1\quad \text{for$1\lt x\lt 2$}\end{cases}$ is correct. It is a mixed discrete / continuous density. – wolfies Nov 4 '13 at 4:46
• @wolfies: it is completely incorrect. What this has at $x=1$ is a point mass, not a density. A way to indicate a point mass might be with a Dirac delta, but certainly not a density value of $1/2$ at a point. – Robert Israel Nov 4 '13 at 5:38