IAI Actuarial Core Principles · Actuarial Statistics · Jointly distributed random variables
X is a Bernoulli indicator with P(X=1)=0.5. Given X=0, Y is uniform on (0,10); given X=1, Y is uniform on (0,20). What is P(X=1 | Y=15)?
The probability is 1. A value of Y equal to 15 has zero density when X is 0 because that uniform lives on (0,10), while it has positive density when X is 1. So the observation can only have come from X=1.
- A0.5
- B1/3
- C2/3
- D1Correct
- 0.75
Explanation
Y=15 lies outside the support (0,10) when X=0, so its conditional density there is 0. Thus the joint density is positive only for X=1, giving probability 1. Options such as 2/3 wrongly compare densities as if both were possible.
Did you get it right without looking?
One question tells you little. A timed set on Jointly distributed random variables shows your real accuracy, how long you take and where you lose marks.
More Jointly distributed random variables questions
- N is Bernoulli with P(N=1)=0.3. Given N=1, X has density 2e^(-2x) for x>0; given N=0, X has density e^(-x) for x>0. What is P(N=1 | X>1)?
- X has a Poisson distribution with mean 3. Given X=x, Y is binomial with x trials and success probability 0.4. What is E[Y]?
- X is uniform on (-1, 1) and Y = X^2. Which statement about X and Y is correct?
- X and Y are independent and each uniformly distributed on (0,1). Let S = X + Y. What is P(S > 1.5)?
- The joint probability function of X and Y is P(X=0,Y=0)=0.1, P(X=0,Y=1)=0.3, P(X=1,Y=0)=0.2 and P(X=1,Y=1)=0.4. What is E[Y | X=1]?
- X and Y have joint density f(x,y) = 2 for 0 < y < x < 1, and zero elsewhere. What is E(Y | X = 0.6)?