Suppose that random variable U follows a continuous Uniform distribution with parameters 0 and 10 (i.e. U∼U(0,10) )

Now let’s denote A the event that U = 5 and B the event that U is equal either to 5 or 6.

According to my understanding, both events have zero probability to occur.Now, if we consider to compute P(A|B) , we cannot use the conditional law P(A|B)=P(A∩B)P(B), because P(B) is equal to zero.

However, my intuition tells me that P(A|B)=1/2.

**Answer**

“The concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible.” A. Kolmogorov

For continuous random variables, X and Y say, conditional distributions are defined by the property that they recover the original probability measure, that is, for all measurable sets A∈A(X), B∈B(Y),P(X∈A,Y∈B)=∫BdPY(y)∫BdPX|Y(x|y)This implies that the conditional density is defined arbitrarily on sets of measure zero or, on other words, that the conditional density pX|Y(x|y) is defined *almost everywhere*. Since the set {5,6} is of measure zero against the Lebesgue measure, this means that you can define both p(5) and p(6) in absolutely arbitrary manners and hence that the probability P(U=5|U∈{5,6})can take any value.

This does not mean you cannot define a conditional density by the ratio formula f(y|x)=f(x,y)/f(x)as in the bivariate normal case but simply that the density is only defined almost everywhere for both x and y.

“Many quite futile arguments have raged – between otherwise competent

probabilists – over which of these results is ‘correct’.” E.T. Jaynes

The fact that the limiting argument (when ϵ goes to zero) in the above answer seems to give a natural and intuitive answer is related with Borel’s paradox. The choice of the parametrisation in the limit matters, as shown by the following example I use in my undergrad classes.

*Take the bivariate normal X,Yi.i.d.∼N(0,1) What is the conditional density of X given that X=Y?*

If one starts from the joint density φ(x)φ(y), the “intuitive” answer is [proportional to] φ(x)2. This can be obtained by considering the change of variable (x,t)=(x,y−x)∼φ(x)φ(t+x) where T=Y−X has the density φ(t/√2)/√2. Hence f(x|t)=φ(x)φ(t+x)φ(t/√2)/√2 and f(x|t=0)=φ(x)φ(x)φ(0/√2)/√2=φ(x)2√2 However, if one considers instead the change of variable (x,r)=(x,y/x)∼φ(x)φ(rx)|x| the marginal density of R=Y/X is the Cauchy density ψ(r)=1/π{1+r2} and the conditional density of X given R is f(x|r)=φ(x)φ(rx)|x|×π{1+r2} Therefore, f(x|r=1)=πφ(x)2|x|/2.

And here lies the “paradox”: the events R=1 and T=0 are the same as X=Y, but they lead to different conditional densities on X.

**Attribution***Source : Link , Question Author : Noob , Answer Author : Xi’an*