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I have the CDF: $$ F(x)= \begin{cases} 0 & \text{if } x < 1 \\ \frac{x^2-2x+2}{2} & \text{if } 1 \le x < 2 \\ 1 & \text{if } x \ge 2 \end{cases} $$ I want to find the PDF and I noticed that $F$ is not continuous (at $x=1$), but it is right continuous. Therefore it's still a valid CDF. And I know, after reading different posts on this site, that the PDF is $$ f(x)= \begin{cases} 0 & \text{if } x < 1 \\ \frac{1}{2} & \text{if } x=1 \\ x-1 & \text{if } 1 < x < 2 \\ 0 & \text{if } x \ge 2 \end{cases} $$ My question is this: Why (according to what definition/fact/theorem) is the probability mass $\frac{1}{2}$ at $x=1$?

If you blindly differentiate the CDF, piece-by-piece, you lose that information; at least I did. (Edit) I always thought the probability of a single point from a continuous random variable was $0$. (End edit)

Thank you in advance for your help and insights.

user78000
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    As $x$ approaches $1$ from the right, the cdf approaches $1/2$. As we approach from the left, the cdf approaches $0$. So there must be a point mass of $1/2$ at $x=1$. – André Nicolas Jun 15 '13 at 03:03
  • By the way, I would not say the pdf is $1/2$ at $1$. – André Nicolas Jun 15 '13 at 03:08
  • I recognize that question from the Society of Actuaries' Exam P sample questions. If you are studying for an actuarial exam, one thing you should be cautious about is when the SOA gives you a CDF and asked to calculate something using that. – Clarinetist Jun 15 '13 at 03:19
  • @Clarinetist What do you mean? Do they give tricky CDFs? – user78000 Jun 15 '13 at 03:30
  • @user78000 - Yes, I can say this from experience. One of the CDFs I was given I didn't know how to use to solve the problem. – Clarinetist Jun 15 '13 at 03:32
  • @AndréNicolas Then how would you phrase what's happening with the PDF at $x=1$? Given a "small" interval about $x=1$, the PDF is $\frac{1}{2}$? – user78000 Jun 15 '13 at 03:36
  • I would not even use the term pdf. There is indeed density for $x\gt 1$. But the distribution is "mixed." – André Nicolas Jun 15 '13 at 03:38

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Use the fact that $\int f(x) dx = 1$ to identify what the pdf must be at $x=1$. Once you check this condition you realize that the pdf must have a mass point of $\frac{1}{2}$ at $x=1$.

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  • So is there some definition out there that looks something like the following: If $F$ is a CDF (strictly right continuous) with a jump discontinuity at $x=a$, and $f(x)$ is the corresponding PDF, then $f(a)=F(a)$? – user78000 Jun 15 '13 at 03:11
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Ah, this looks like an actuarial science problem I've seen. Yes, when you do differentiate, you do lose the point mass. Notice something though - if $F'(x)$ directly equaled the pdf, that would mean that $\int\limits_{1}^{2}(x-1) \text{ d}x = 1$, but clearly, this isn't the case since this integral leads to $2^2/2 - 2 - (1/2 - 1) = 1/2$. Notice, though, that $F(1) = \dfrac{1}{2}$. This implies that $P(X \leq 1) = \dfrac{1}{2}$, so there must be something that is adding a probability at $x = 1$, which implies your $f(x)$.

In case you want to read up more on this concept, $X$ is said to follow a mixed distribution.

Clarinetist
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    Yes yes! Indeed, it is an actuary problem. I'm studying for Exam P. My studies have been going very well as most of the concepts needed to solve the problems I have seen are review for me. This is allowing me to dig much deeper into topics where if I were to have less time and knowledge, I would simply rote memorize and move on. Thanks for the help. – user78000 Jun 15 '13 at 03:20
  • If you would like any help (I just passed Exam P in March with an 8), please don't hesitate to ask around here. I enjoy seeing actuarial questions on this site. – Clarinetist Jun 15 '13 at 03:21