2

Probability distribution function (pdf): \begin{align*} &P(-\infty<X<+\infty)=1\\ &P(x_0\leq X\leq x_1)\geq 0\\ &p(x_0)=-9.0? \end{align*}

Source.

Why does the third one make sense? I know that $p(x)$ cannot be negative, yet supposedly the third one can occur. Why is that?

Em.
  • 15,981
Armon Safai
  • 263
  • Where does that come from? – Clement C. Jan 20 '16 at 03:06
  • @ClementC. My machine learning professor – Armon Safai Jan 20 '16 at 03:12
  • Well.. how did he define a pdf? That's where you should start. For instance, if you take the pdf of a Gaussian random variable and modify it on one single point, say $x_0=0$, to put the value $-9.0$ at that single point -- does that still satisfy the definition of a probability density function? – Clement C. Jan 20 '16 at 03:16
  • @ClementC. the general way a pdf is defined. Well that still violates the definition of pdf cause the p(x) function cannot be negative on any point. – Armon Safai Jan 20 '16 at 03:18
  • Why is that? Again, to start answering that question, you need to start with a rigorous definition of what a pdf is... if you don't have one, then how can you argue what violates it? – Clement C. Jan 20 '16 at 03:18
  • See http://math.stackexchange.com/questions/580817/can-a-probability-density-function-take-negative-values – Clement C. Jan 20 '16 at 14:39

1 Answers1

0

The third one can definitely not occur, whether it's discrete or continuous, because it's negative. I'm not sure where you're hearing that's valid.

Clarinetist
  • 19,519
  • What definition are you basing the above answer on? Right now, it's completely arbitrary... – Clement C. Jan 20 '16 at 03:14
  • @ClementC. No matter your probability measure $P$, is it not true that $P \geq 0$? – Clarinetist Jan 20 '16 at 03:36
  • @ClementC. This is not "arbitrary," it's by the mere definition of a probability measure, whether that's a CDF or a PDF or a probability mass function or whatever have you. – Clarinetist Jan 20 '16 at 03:37
  • Please, define what you start with. You haven't defined what a pdf is, neither even what a cdf is. You just base your answer on a "well-known definition of a pdf," but that very definition is exactly what the OP's question is about. Probabilities are non-negative, but a probability density function is not a probability. Would you say that the probability to have $X=1/2$, when X is uniform in $[0,1]$, is its density at the point, i.e. $1$? – Clement C. Jan 20 '16 at 13:27
  • Also, see http://math.stackexchange.com/questions/580817/can-a-probability-density-function-take-negative-values – Clement C. Jan 20 '16 at 14:01
  • @ClementC. This is getting to the point of nitpicking. If you've taken measure theory, you would know that the CDF is a probability measure of an interval. Probability measures are defined to be $\geq 0$. The PMF, as well, is a probability measure. Furthermore, if $F_X$ is a CDF, any function $f_X$ satisfying $$F_X(x) = \int_{-\infty}^{x}f_X(t)\text{ d}t$$ for all $x$ is a PDF. Since $F_X$ is a probability measure (no matter the value of $x$ above), it follows that $f_X \geq 0$ as well. – Clarinetist Jan 20 '16 at 14:35
  • Exactly no. You are saying "any function $f_X$ satisfying this [...]" -- and these functions need not be non-negative. They only need to be non-negative almost everywhere. Which the question is about... a point has Lebesgue measure zero. – Clement C. Jan 20 '16 at 14:37
  • @ClementC. Okay, sure, outside of a set of measure zero. But I think it's reasonable to believe that someone who is just being introduced to what a CDF is isn't going to worry about these technicalities. But no matter, I can't delete the answer anyway. – Clarinetist Jan 20 '16 at 14:38
  • @ClementC. If OP was told that $p$ cannot be negative, it is reasonable to assume that (s)he is likely taking an undergrad class in probability, where the variables are either absolutely continuous or simple. Furthermore, judging by past question history, they recently asked about counting methods in probability. It would be inappropriate to be bringing in a.e. issues in such a class. – Clarinetist Jan 20 '16 at 14:48