Let X and Y be independent continuous random variables that are uniformly distributed on (0,1). Let H=(X+2)Y. Find the probability \mathbf{P}(\ln H\geq z) where z is a given number that satisfies \displaystyle e^ z<2. Your answer should be a function of z.

Question

Hint: Condition on X.

\mathbf{P}(\ln H\geq z)=\qua

Answer 1

Let's condition on the value of X.

If X is in the interval (0,1/2), then (X+2) will be in the interval (2, 5/2). In this case, \ln H will have the range (\ln 2, \ln(5/2)).

If X is in the interval (1/2,1), then (X+2) will be in the interval (5/2, 3). In this case, \ln H will have the range (\ln(5/2), \ln(3)).

Since X is uniformly distributed on (0,1), the probability that X is in the interval (0,1/2) is 1/2, and the probability that X is in the interval (1/2,1) is also 1/2.

Therefore, the probability that \ln H \geq z is equal to:

(1/2) * P(\ln H \geq z | X \in (0,1/2)) + (1/2)*P(\ln H \geq z | X \in (1/2,1)).

For X in the interval (0,1/2), the probability that \ln H \geq z is equal to P(\ln H \geq z | X \in (0,1/2)) = P(\ln((X+2)Y) \geq z | X \in (0,1/2)) = P(Y \geq e^z/(X+2) | X \in (0,1/2)).

Since X and Y are independent, this probability is equal to P(Y \geq e^z/(X+2)).

Similarly, for X in the interval (1/2,1), the probability that \ln H \geq z is equal to P(Y \geq e^z/(X+2)).

Therefore, the probability that \ln H \geq z is:

(1/2) * P(Y \geq e^z/(X+2) | X \in (0,1/2)) + (1/2) * P(Y \geq e^z/(X+2) | X \in (1/2,1)).

Now, since Y is uniformly distributed on (0,1), the probability that Y \geq e^z/(X+2) is equal to 1 - e^z/(X+2).

Therefore, the probability that \ln H \geq z is:

(1/2) * (1 - e^z/(X+2)) | X \in (0,1/2) + (1/2) * (1 - e^z/(X+2)) | X \in (1/2,1).

Simplifying further, we get:

(1/2) * (1 - e^z/(X+2)) + (1/2) * (1 - e^z/(X+2)).

= 1 - e^z/(X+2).

Therefore, the probability \mathbf{P}(\ln H\geq z) is given by the function 1 - e^z/(X+2).