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)=\quad
Let X be a standard normal random variable, and let F_ X(x) be its CDF. Consider the random variable Z=F_ X(X). Find the PDF f_ Z(z) of Z. Note that f_ Z(z) takes values in (0,1).
f_ Z(z)=\quad
We can start by finding the probability distribution function (PDF) of H.
Let's first condition on X. Since X is uniformly distributed on (0,1), the conditional PDF of X given a specific value x is 1 for 0 < x < 1 and 0 otherwise.
Next, we can consider the random variable Y. Since Y is also uniformly distributed on (0,1), its PDF is also 1 for 0 < y < 1 and 0 otherwise.
Now, let's find the PDF of H given a specific value x:
Since H = (X+2)Y, we can rewrite this as H = XY + 2Y.
The PDF of XY can be found by convolution:
f_{XY}(h|x) = ∫ f_X(x') f_Y(h/x') dx'
Since X and Y are independent, their joint PDF is simply the product of their individual PDFs:
f_{XY}(h|x) = f_X(x) f_Y(h/x)
Since both X and Y are uniformly distributed on (0,1), their PDFs are both 1 for 0 < x,y < 1. Therefore, the joint PDF simplifies to:
f_{XY}(h|x) = 1
Now, let's consider the random variable 2Y. The PDF of 2Y can be found by a change of variables:
f_{2Y}(h|x) = f_Y(h/2)
Again, since Y is uniformly distributed on (0,1), its PDF is 1 for 0 < y < 1. Therefore, the PDF of 2Y is:
f_{2Y}(h|x) = 1
Finally, to find the PDF of H, we need to convolve the PDFs of XY and 2Y:
f_H(h|x) = ∫ f_{XY}(h-h') f_{2Y}(h') dh'
Since both PDFs are constant (equal to 1), the convolution simplifies to:
f_H(h|x) = area of overlap of [0,1] with [h-1,h] = min(h,1) - max(h-1,0)
In other words, the PDF of H given X = x is:
f_H(h|x) = min(h,1) - max(h-1,0)
Now, let's find the cumulative distribution function (CDF) of H:
F_H(h|x) = ∫ f_H(t|x) dt for t from -∞ to h
= ∫ (min(t,1) - max(t-1,0)) dt for t from -∞ to h
= [t^2/2 - (t-1)^2/2] from -∞ to h
= (h^2/2 - (h-1)^2/2) - (0^2/2 - (-1)^2/2)
= (h^2/2 - (h^2 - 2h + 1)/2) - (0 - 1/2)
= (h^2 - h^2 + 2h - 1)/2 + 1/2
= (2h - 1)/2 + 1/2
= h/2 + 1/2
Therefore, the CDF of H given X = x is:
F_H(h|x) = h/2 + 1/2
Now, let's find the CDF of H by conditioning on X:
F_H(h) = ∫ F_H(h|x) f_X(x) dx for x from 0 to 1
= ∫ (h/2 + 1/2) dx for x from 0 to 1
= (hx/2 + x/2) from 0 to 1
= (h/2 + 1/2) - (0/2 + 0/2)
= h/2 + 1/2
Therefore, the CDF of H is:
F_H(h) = h/2 + 1/2
Now, let's find the probability \mathbf{P}(\ln H\geq z):
\mathbf{P}(\ln H\geq z) = \mathbf{P}(H\geq e^z)
Since the CDF of H is F_H(h) = h/2 + 1/2, we can find the probability as:
\mathbf{P}(\ln H\geq z) = 1 - \mathbf{P}(H < e^z)
= 1 - F_H(e^z)
= 1 - (e^z/2 + 1/2)
= 1/2 - e^z/2
Therefore, the probability \mathbf{P}(\ln H\geq z) is 1/2 - e^z/2.
To find the probability \(\mathbf{P}(\ln H \geq z)\), we can make use of the given hint to condition on X.
Let's start by finding the conditional probability \(\mathbf{P}(\ln H \geq z | X = x)\) for a fixed value of x.
We have H = (X+2)Y, where X and Y are independent continuous random variables that are uniformly distributed on (0,1).
Given X = x, we can rewrite H as H = (x+2)Y.
Now, we want to find the probability \(\mathbf{P}(\ln H \geq z | X = x)\).
Since \(\ln(\cdot)\) is a monotonically increasing function, we can rewrite the inequality as H \(\geq\) e^z.
Substituting H = (x+2)Y, we have (x+2)Y \(\geq\) e^z.
Since Y is uniformly distributed on (0,1), its PDF is 1 on the interval (0,1) and 0 otherwise.
For the case when (x+2) is positive, the inequality is satisfied if and only if Y \(\geq\) e^z / (x+2).
The probability of Y \(\geq\) e^z / (x+2) is equal to \(1 - \frac{{e^z}}{{x+2}}\), since the probability of Y falling in the interval (0, e^z / (x+2)) is e^z / (x+2).
For the case when (x+2) is negative, the inequality is always satisfied since Y is positive.
Therefore, the conditional probability \(\mathbf{P}(\ln H \geq z | X = x)\) is given by:
\(\begin{cases} 1 - \frac{{e^z}}{{x+2}}, & \text{if } x+2 > 0 \\ 1, & \text{if } x+2 \leq 0 \end{cases}\)
Now, we need to calculate the probability \(\mathbf{P}(X = x)\) for each x.
Since X is uniformly distributed on (0,1), its PDF is 1 on the interval (0,1) and 0 otherwise.
Therefore, \(\mathbf{P}(X = x) = 1\) for \(0 < x < 1\), and \(\mathbf{P}(X = x) = 0\) otherwise.
Now, we can use the law of total probability to find \(\mathbf{P}(\ln H \geq z)\).
\(\mathbf{P}(\ln H \geq z) = \int_{0}^{1} \mathbf{P}(\ln H \geq z | X = x) \cdot \mathbf{P}(X = x) \,dx\)
For \(0 < x < 1\), the conditional probability \(\mathbf{P}(\ln H \geq z | X = x)\) is given by \(1 - \frac{{e^z}}{{x+2}}\),
and \(\mathbf{P}(X = x) = 1\).
So, the integral simplifies to:
\(\mathbf{P}(\ln H \geq z) = \int_{0}^{1} 1 - \frac{{e^z}}{{x+2}} \,dx = \int_{0}^{1} 1 \,dx - e^z \int_{0}^{1} \frac{1}{{x+2}} \,dx\)
Using the property \(\int \frac{1}{{a+x}} \,dx = \ln(a+x) + C\), where C is a constant of integration,
we can evaluate the second integral:
\(\int_{0}^{1} \frac{1}{{x+2}} \,dx = \ln(x+2) \Big|_{0}^{1} = \ln(3) - \ln(2) = \ln(3) - \ln(2)\)
Therefore, the probability \(\mathbf{P}(\ln H \geq z)\) is given by:
\(\mathbf{P}(\ln H \geq z) = \int_{0}^{1} 1 \,dx - e^z \cdot (\ln(3) - \ln(2)) = 1 - e^z \cdot (\ln(3) - \ln(2))\)
Finally, we can write the answer as a function of z:
\(\mathbf{P}(\ln H \geq z) = 1 - (\ln(3) - \ln(2)) \cdot e^z\)