Conditional probability of continuous variable

Suppose that random variable U follows a continuous Uniform distribution with parameters 0 and 10 (i.e. UU(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(AB)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 AA(X), BB(Y),P(XA,YB)=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,yx)φ(x)φ(t+x) where T=YX 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)22 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

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