Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need somebody to help me to understand the following concepts:

Assuming $X$, $Y$ are random variables (r.v.'s). What does the following represents:

  1. $P(X+Y|Y)$, what is this?
  2. $P(X+Y|Y=y)$, my understanding it is a r.v..
  3. $P(X+Y=s|Y)$,my understanding it is a r.v..
  4. $P(X+Y=s|Y=y)$, my understanding it is a number.

How do they relate to each other?

Also, in case of Y a continuous r.v., consider $P(X+Y|Y=y)$, but $P(Y=y)$ is always zero. How can this conditioning be thought of?

Thanks a lot.


I haven't seen people giving answers to my question: in case of $Y$ a continuous r.v., consider $P(X+Y|Y=y)$, but $P(Y=y)$ is always zero. How can this conditioning be thought of? Just to give another example (in addition to the one I gave in comments about the uniform distribution), consider standard Brownian motion. $Pr(B_t\ge a|B_s=b)$ is clearly sensible and different from $Pr(B_t\ge a)$ for $t\ge s$. But here $Pr(B_s=b)$ is zero.

Also I read in books on "Markov Chains", for example, the notation of Markov property is stated as: $Pr(C_t|C_{t-1}, ..., C_1)=Pr(C_t|C_{t-1})$ So there is the notation $P(X)$ or for that matter, $P(X|Y)$. Is this notation short for $P(X=\text{any value}|Y=\text{any value})$?

share|improve this question
Are you using the + sign to denote addition or the logical 'and'? –  Chris Taylor Dec 15 '11 at 21:53
No, actually people do use the notation. Think about the example when $X$ and $Y$ are uniform in interval $[0,1]$, $P(X+Y \ge 0.4|y=0.1)=P(X \ge 0.3)$ –  littleEinstein Dec 15 '11 at 21:55
@ChrisTaylor: no, '+' is just the normal addition. r.v. $X$ and $Y$ can form a new r.v. $X+Y$. –  littleEinstein Dec 15 '11 at 22:12
I've seen "$+$" used for logical "or"; I don't think I've ever seen it used for "and". –  Michael Hardy Dec 16 '11 at 0:45
I've seen my students use '+' to denote 'and' - hence the question! –  Chris Taylor Dec 21 '11 at 17:26

2 Answers 2

In probability, we call outcomes $\omega$ (omega), and the set of all outcomes $\Omega$ (big omega). $P$, a probability measure, is a function that maps subsets of $\Omega$ into [0,1]. It must always be a function from a set into [0,1]. A random variable is a function on $\Omega$, not a subset of it, and what we really mean by $P(X)$, then, must be defined more clearly.

Probability is defined over events, not random variables. So rather than saying $P(X)$, you should be thinking of things like $P($coin lands heads$)$ or $P($it rains tomorrow$)$. In this case, $X$ refers to either a coin toss, or the weather, and if we ever write $P(X)$, it's really just short for $P(X=x)$

But $X=x$ is not a subset of $\Omega$ either, and what we really mean is $$P(\{ \omega \in \Omega |X(\omega)=x\}),$$ the probability measure of all events $\omega$ in which the function $X$ maps $\omega$ into $x$. Since that is quite a long thing to write, people refer to it as $P(X)$ for short, understanding that it may take several different values depending on which value $X$ takes. $X$, in turn, depends on which value $\omega$ takes. Now to conditioning.

Your intuition is correct. $P(X=x|Y)$ is a random variable, and $P(X=x|Y=y)$ is a number (adding $Y$ to $X$ doesn't really change anything).
So consider tossing a die, then we can call an outcome a number between 1 and 6; that is, $\omega_1=1,\omega_2=2...\omega_6=6$, and no others. If it's a fair die, then we'll have $P(\omega_i)=1/6$. For example, we can have we can have $X=1$ if $\omega$ is even, and $X=0$ otherwise. Or $Y=2$ if $\omega$ is prime, $Y=1$ if $\omega$ is composite, and $Y=0$ otherwise. By definition
$P(X=x):=P($set of all $\omega$ which make $X=x)$, and
$P(X=x|Y=y):=P($set of all $\omega$ which make $X=x$ and $Y=y$)/$P(Y=y)$.

$$P(X=1|Y=2)=\frac{P(\{2,4,6\}\cap\{2,3,5\})}{P(\{2,3,5\})}=1/3$$ $$P(X=1|Y=1)=\frac{P(\{2,4,6\}\cap\{4,6\})}{P(\{4,6\})}=1$$ $$P(X=1|Y=0)=\frac{P(\{2,4,6\}\cap\{1\})}{P(\{1\})}=0$$
As you can see, $W=P(X=x|Y)$ is a random variable depending on what $Y$ equals, and we have $$P(W=1/3)=P(Y=2)=P\{2,3,5\}=1/2$$ $$P(W=1)=P(Y=1)=P(\{4,6\}=1/3$$ $$P(W=0)=P(Y=0)=P(\{1\})=1/6$$ As you can calculate, $$\mathbb{E}(W) = \sum(W P(W)) = P(X=x)$$

That is to say, the expectation of the conditional probability of $X$ is just the probability of $X$, for each value that $X$ takes.

share|improve this answer
The things called $\omega$ are not "events"; they're "outcomes". An event is a set of outcomes. –  Michael Hardy Dec 16 '11 at 1:56
@MichaelHardy: Did you see my edit? What is your thought on that? –  littleEinstein Dec 20 '11 at 20:40
@jalopezp: Did you see my edit? What is your thought on that? –  littleEinstein Dec 20 '11 at 21:01

On the basis that $X$ and $Y$ are numerical random variables with known distributions, and $X$ takes discrete values, then both (1) and (2) are shorthands for (3) and (4) that are potentially confusing and probably meaningless when considered carefully. Even $P(X)$ on its own is not particularly meaningful.

For (3) , if it is not a shorthand for (4), you are correct that $P(X+Y=s|Y)$ is a random variable if $s$ is known: it is a function of $Y$, which is a random variable. It is a random function of $s$ when $s$ is unknown.

For (4) you are also correct that $P(X+Y=s|Y=y)$ is a number if $s$ and $y$ are known: it is a probability so must be in the interval $[0,1]$. It is a function of $s$ and $y$ when they are unknown.

share|improve this answer
Did you see my edit? What is your thought on that? –  littleEinstein Dec 20 '11 at 20:40
The first part of your addition about conditioning on a particular value of a continuous distribution is an important but different question. The second part of your question is about another shorthand: $\Pr(C_t|C_{t-1}, \ldots, C_1)=\Pr(C_t|C_{t-1})$ is saying $\Pr(C_t=c_t|C_{t-1}=c_{t-1}, \ldots, C_1=c_1)=\Pr(C_t=c_t|C_{t-1}=c_{t-1})$ for all possible $c_i$ in the supports of the $C_i$. –  Henry Dec 20 '11 at 21:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.