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.
Hint: Condition on X.
\mathbf{P}(\ln H\geq z)=\qua
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).
To find the probability \(\mathbf{P}(\ln H \geq z)\), we can first consider the probability \(\mathbf{P}(\ln H \geq z | X = x)\) for a given value of \(x\) for the random variable \(X\).
Since \(X\) and \(Y\) are independent continuous random variables uniformly distributed on (0, 1), the joint probability density function can be written as \(f_{XY}(x,y) = 1\) for \(0 < x < 1\) and \(0 < y < 1\).
Now, let's find the cumulative distribution function (CDF) of the random variable \(H\) when \(X = x\).
We have \(H = (X + 2)Y\). Substituting \(X = x\), we get \(H = (x + 2)Y\).
To find the CDF of \(H\), denoted as \(F_H(h)\), we compute:
\[
\begin{align*}
F_H(h) &= \mathbf{P}(H \leq h) \\
&= \mathbf{P}((x + 2)Y \leq h) \\
&= \mathbf{P}\left(Y \leq \frac{h}{x + 2}\right).
\end{align*}
\]
Since \(Y\) follows a uniform distribution on (0, 1), we can write:
\[
F_H(h) = \begin{cases}
0 & \text{ for } h < 0 \\
\frac{h}{x + 2} & \text{ for } 0 \leq h < (x + 2) \\
1 & \text{ for } h \geq (x + 2).
\end{cases}
\]
Now, let's condition on \(X\) to find the probability \(\mathbf{P}(\ln H \geq z | X = x)\).
\[
\mathbf{P}(\ln H \geq z | X = x) = \mathbf{P}\left(H \geq e^z \middle| X = x\right) = 1 - F_H(e^z).
\]
Substituting the expression for \(F_H(h)\) into the above equation, we get:
\[
\mathbf{P}(\ln H \geq z | X = x) = 1 - \begin{cases}
0 & \text{ for } e^z < 0 \\
\frac{e^z}{x + 2} & \text{ for } 0 \leq e^z < (x + 2) \\
1 & \text{ for } e^z \geq (x + 2).
\end{cases}
\]
Finally, to find the probability \(\mathbf{P}(\ln H \geq z)\), we need to consider the possible values of \(X\) and integrate over its range (0, 1). Since \(X\) is uniformly distributed on (0, 1), its probability density function is given by \(f_X(x) = 1\) when \(0 < x < 1\).
Using the law of total probability, we have:
\[
\mathbf{P}(\ln H \geq z) = \int_0^1 \mathbf{P}(\ln H \geq z | X = x) f_X(x) dx.
\]
Plugging in the expression for \(\mathbf{P}(\ln H \geq z | X = x)\), we get:
\[
\mathbf{P}(\ln H \geq z) = \int_0^1 1 - \begin{cases}
0 & \text{ for } e^z < 0 \\
\frac{e^z}{x + 2} & \text{ for } 0 \leq e^z < (x + 2) \\
1 & \text{ for } e^z \geq (x + 2).
\end{cases} dx.
\]
Simplifying this integral will yield the final expression for the probability \(\mathbf{P}(\ln H \geq z)\) as a function of \(z\).