# Joint probability of iid samples where pdf is not continuous at some points

Suppose $y_i$ is a random variable generated by sampling $y_i' \sim \mathcal{N}(0,1)$ and setting $y_i = y_i'$ if $c_1 \leq y_i' \leq c_2$, $y_i = c_1$ if $y_i' < c_1$, and $y_i = c_2$ if $y_i' > c_2$. In other words, take a normally distributed sample $y_i'$ and threshold it at some lower and upper limit to give $y_i$. The "pdf" of $y_i$ then looks like a normal distribution for $c_1 < y_i < c_2$, and is $0$ everywhere else except at $c_1,c_2$ where it is some other value.

Let $Y_j \in \mathbb{R}^T$ be a vector of iid samples $(y_1, \ldots ,y_T)$ where each $y_i$ is distributed in this way, and suppose I have a set of vectors $\{Y_j\}, j = 1, \ldots ,N$.

I want to calculate $p(Y_j)$ = $p(y_{j,1},\ldots ,y_{j,T})$ for each $Y_j$ in a way that I can compare this value for different $Y_j$ in a meaningful way. If there was no discontinuity in the pdf of $y_i$ this would be easy - just multiply the pdf of the iid components: $$p(Y_j) = \prod_{i=1}^T p(y_{j,i})$$

I'm confused as to what to do when the pdf of each $y_i$ is not continuous. If I work it out as a product like this, what value do I use for $p(y_i)$ if $y_i = c_1$ or $c_2$? If I use for instance: $$p(c_2) = \int_{c_2}^{\infty} \mathcal{N}(y_i; 0,1) \mathbb{d} y_i$$

This integral could be much lower than $p(c_2 - \epsilon)$ even though the "probability of seeing $c_2$" is much higher than a value slightly less than $c_2$. When I take the product I think the value for $p(y_i = c_2)$ should be larger than for $p(y_i = c_2 - \epsilon)$, otherwise comparing different $p(Y_j$) is meaningless.

My actual situation is that I have a set of measurements $Y_j$ in a file, and need to calculate $p(Y_j)$ for each.

Not really an answer but worth mentioning:

The rv $y_i$ as described in top of your question has no PDF.

If it would have - let's call if $f$ - then $P(y_i\in A)=\int_Af(y)dy$ should be true for each measurable set $A$.

However, if $\emptyset\neq A\subseteq\{c_1,c_2\}$ then the LHS is positive and the RHS is zero.

• Seems like an answer to me, probably even the answer. – Did Oct 3 '14 at 8:41
• What does this mean for $P(Y_j)$? – akxlr Oct 3 '14 at 8:47
• To avoid misunderstandings: what exactly is meant by $P(Y_j)$ (or $p(Y_j)$ in the question)? You are saying that you want to calculate it without mentioning what it is. – drhab Oct 3 '14 at 8:53
• I mean it to be the probability of observing the set of $T$ measurements $y_{j,1}, \ldots ,y_{j,T}$ - i.e. each $Y_j$ is a vector of $T$ iid measurements and I want to work out the probability of observing all of them assuming they are independent. – akxlr Oct 3 '14 at 9:13
• Probability of "observing..." sounds vague. Its probability can only be calculated if "observing..." is some sort of event, i.e. a measurable subset of a probability space. Can you describe this event in clear terms of probability theory? – drhab Oct 3 '14 at 9:23