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Conditioning is a tool I have used a lot in the discrete setting: usually this takes the form $$P(X=k) = \sum_{i=0}^\infty P(X=k | Y=i) P(Y=i).$$ I'm a little confused about the analogous situation for continuous random variables.

Suppose we have this little problem, which I am making up for illustration purposes:

Let $X$ be normally distributed: $X \sim N(0,1)$. Now $Y$ will be another random variable that is $N(1,1)$ if the outcome of $X$ is larger than $2$, and is $N(-1,1)$ if the outcome of $X$ is negative. If $X \in [0,2]$ then $Y$ simply takes the same value of $X$.

What is $P(Y>0)$?

Obviously $Y$ depends on $X$, but I get mixed up when it comes down to trying to formulate the conditioning relationship. I don't really care about the numberical answer here, but would like to see how the cumulative and density functions should be set up in order to calculate the answer.

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  • $\begingroup$ For the scenario you set up X and Y could be independent or dependent. If independent the probabilities multiply regardless of conditioning. Otherwise the joint density depends the correlation coefficient $\rho$.between X and Y. $\endgroup$
    – Michael R. Chernick
    Commented Apr 1, 2017 at 5:08
  • $\begingroup$ Conditioning for continuous densities is essentially the same as for discrete densities, but with summation replaced by integration over the full domain. There are some complexities that occur when the events you are conditioning on have zero measure, but that doesn't occur in your example. Try working through the example treating integrals as you would summations. $\endgroup$
    – AaronDefazio
    Commented Apr 1, 2017 at 6:14
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    $\begingroup$ What if $X \in [0,2]$ and so is neither larger than 2 nor less than 0, What is the distribution of $Y$ in this case? Please do not say that $Y$ is undefined in this case, because that is an event of positive probability and $Y$ must have some value when this event occurs, even if it is 0. $\endgroup$
    – Dilip Sarwate
    Commented Apr 1, 2017 at 15:02
  • $\begingroup$ I have edited the question to include the outcome $X \in [0,2]$. $\endgroup$
    – theQman
    Commented Apr 2, 2017 at 2:24
  • $\begingroup$ @AaronDefazio what throws me off is taking the probability of a single point outcome. In the discrete case we have $P(Y=i)$, but this would be $0$ in a continuous setting. $\endgroup$
    – theQman
    Commented Apr 2, 2017 at 2:27

1 Answer 1

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The question "What is $P\{Y > 0\}$?" has a very simple answer that follows the same kind of calculations as in the OP's calculations of the probability mass function of a discrete random variable via the law of total probability.

Since $X \sim N(0,1)$, we have three events: $A = \{-\infty < X < 0\}$, $B = \{0 \leq X \leq 2\}$, and $C=\{2 < X \leq \infty\}$ with probabilities $P(A) = \Phi(0) = \frac 12$, $P(B) = \Phi(2)-\Phi(0) = \Phi(2)-\frac 12$, and $P(C) = 1-\Phi(2)$. Then, $$P\{Y > 0\} = P\{Y > 0\mid A\}P(A) + P\{Y > 0\mid B\}P(B) + P\{Y > 0\mid C\}P(C). \tag{1}$$ But,

  • Given that event $A$ has occurred, the conditional distribution of $Y$ given $A$ is $N(-1,1)$ and thus $P\{Y > 0\mid A\} = 1-\Phi(1)$ where $\Phi(\cdot)$ is the standard normal CDF.

  • Similarly, given that event $C$ has occurred, the conditional distribution of $Y$ given $C$ is $N(1,1)$ and thus $P\{Y > 0\mid C\} = 1-\Phi(-1) = \Phi(1)$.

  • Given that event $B = \{0 \leq X \leq 2\}$ has occurred, the value of $Y$ is the same as the value of $X$, and so $Y \in [0,2]$ also. Hence, $P\{Y >0 \mid B\} = 1$.

Substituting into $(1)$ gives \begin{align}P\{Y > 0\} &= \left(1-\Phi(1)\left)\cdot\frac 12\right.\right.+ 1\cdot \left(\Phi(2) - \frac 12\right) + \Phi(1)\cdot(1-\Phi(2))\\ &= \Phi(2) + \frac 12\Phi(1) - \Phi(1)\Phi(2).\tag{2} \end{align}

A small generalization of the above calculation gives the unconditional complementary CDF of $Y$. Instead of $(1)$, consider $$P\{Y > y\} = P\{Y > y\mid A\}P(A) + P\{Y > y\mid B\}P(B) + P\{Y > y\mid C\}P(C) \tag{3}$$ in which it is easy to get that $$P\{Y > y\mid A\} = 1 - \Phi(y+1), \quad P\{Y > y\mid C\} = 1 - \Phi(y-1)$$ but finding $P\{Y > y\mid B\}$ is just a tad trickier. We have that

\begin{align}P\{Y > y\mid B\} &= P\{X >y \mid 0 \leq X \leq 2\}\\ &= \frac{P(\{X > y\}\cap\{0 \leq X \leq 2\})}{P\{0 \leq X \leq 2\}}\\ &= \begin{cases}\displaystyle\frac{P\{0 \leq X \leq 2\}}{P\{0 \leq X \leq 2\}}, & y < 0,\\ \displaystyle\frac{P\{y \leq X \leq 2\}}{P\{0 \leq X \leq 2\}}, & 0 \leq y \leq 2,\\ 0, & y > 2.\end{cases}\\ &= \begin{cases}1, & y < 0,\\ \displaystyle\frac{\Phi(2)-\Phi(y)}{\Phi(2)- \frac 12}, & 0 \leq y \leq 2, \\0, & y > 2.\end{cases} \end{align} Substituting these values into $(3)$ and simplifying is a task left to the OP.

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