# Finding the given conditional distribution

Say a sample $(X_1,X_2,X_3)$ is taken from a $Ber(p)$ distribution, we need to check the conditional distribution of $(X_1,X_2,X_3)$ given a value of $X_1X_2+X_3$, i.e. we need to check $P(X_1=x_1,X_2=x_2,X_3=x_3|X_1X_2+X_3=s)$.

Now, $s$ can take values $0,1,2$.

I am kinda confused here, I am used to tackle conditional probabilities as follows :

I see the 'given' part ($X_1X_2+X_3=s$ here.) as a 'thing' that restricts our sample space. Here our sample space consists of 8 points $(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),(1,1,1)$.

So when we say $P(X_1=0,X_2=0,X_3=0|X_1X_2+X_3=0)$, we say - "First get into the points that satisfy $X_1X_2+X_3=0$, that will be $(0,0,0),(1,0,0),(0,1,0)$, and then check the probability of $(0,0,0)$, which comes out to be $\frac{1}{3}$. "

Now, if we proceed in a 'conventional' manner, i.e. like $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$, the term above comes out to be in the form of $p$ (As given in the solution).

So I wanted to know where does the first method fails ?

Or is it completely wrong ?

• Try substituting $p=\frac 12$ in the “official” answer to see if you get your “simple” answer which assumes that all eight points are equally likely, which is equivalent to assuming that the random variables are $Ber(1/2)$. – Dilip Sarwate May 3 '18 at 11:29
• Make a table of eight rows, one per outcome, in which the first three columns give the values of $X_1,X_2,X_3.$ In a fourth column write the probabilities of each outcome. With simple calculations, taking very little work, you can compute a fifth column giving the values of $X_1X_2+X_3,$ which you might as well label $s$. Focusing on those last two columns, can you now compute the chances $\Pr(s=0),$ $\Pr(s=1),$ and $\Pr(s=2)$? – whuber May 3 '18 at 12:53

I am not sure I understand your question, however I think both methods are ok and give the same result:

P(B) = P((0,0,0),(1,0,0),(0,1,0)) = 3/8

P(A∩B) = P((0,0,0)) = 1/8

so that

P(A∩B)/P(B) = 1/3

In general:

$$P(X_i=0) = 1-p,$$

so

$$P(X_1=0,X_2=0,X_3=0) = (1-p)^3$$

and

$$P(X_1X_2+X_3=0) = (1-p)* [ (1-p)^2+2p*(1-p)]$$

Finally:

$$P(A∩B)/P(B) = \frac{1-p}{1+p}$$

• The solution says something like this: $P(X_1=0,X_2=0,X_3=0 | X_1X_2+X_3=0)=\dfrac{1-p}{1+p}$ – User9523 May 3 '18 at 9:44
• so, if p=1/2, as you have assumed, the solution will be 1/3. – fabiob May 3 '18 at 12:04
• So the 'flaw' I am looking in the first method is that, I am assuming each point is equally likely, right ? – User9523 May 3 '18 at 14:25
• Yes, however let me insist that both methods are valid, if applied correctly. If p=1/2 then you are applying the first method correctly. If p $\neq 1/2$ then, as you are saying, it would be wrong/a flaw to consider that (0,0,0) and (1,0,0) have the same probability to occur. Instead their probabilities to occur would be respectively: $(1-p)^3$ and $p(1-p)^2$. – fabiob May 3 '18 at 15:16