Question

Problem 2. Continuous Random Variables

2 points possible (graded, results hidden)
Let 𝑋 and 𝑌 be independent continuous random variables that are uniformly distributed on (0,1) . Let 𝐻=(𝑋+2)𝑌 . Find the probability 𝐏(ln𝐻≥𝑧) where 𝑧 is a given number that satisfies 𝑒𝑧<2 . Your answer should be a function of 𝑧 .

Hint: Condition on 𝑋 .

1. P(ln H > z) = unanswered

2. Let 𝑋 be a standard normal random variable, and let 𝐹𝑋(𝑥) be its CDF. Consider the random variable 𝑍=𝐹𝑋(𝑋) . Find the PDF 𝑓𝑍(𝑧) of 𝑍 . Note that 𝑓𝑍(𝑧) takes values in (0,1) .

fz(z) = unanswered

Answer 1

[AI-generated answer — Clown Bot]

Why was the math book sad?

Because it had too many problems.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the probability P(lnH > z), we can use the law of total probability and condition on X.

Let's start by finding the distribution of H | X=x. Since X and Y are independent, we have:

H | X=x = (X + 2)Y | X=x = (x + 2)Y

Since X is uniformly distributed on (0, 1), the density function of X is fX(x) = 1 for 0 < x < 1, and fX(x) = 0 otherwise.

Now, let's find the cumulative distribution function of H:

F_H(z | X=x) = P((X + 2)Y ≤ z | X=x)

Since Y is also uniformly distributed on (0, 1), we have:

F_H(z | X=x) = P(Y ≤ z/(x+2)) = z/(x+2) for 0 < z/(x+2) < 1, and F_H(z | X=x) = 1 otherwise.

To find the density function of H, we can differentiate the cumulative distribution function with respect to z:

f_H(z | X=x) = d/dz F_H(z | X=x) = d/dz (z/(x+2)) = 1/(x+2) for 0 < z/(x+2) < 1, and f_H(z | X=x) = 0 otherwise.

Now, let's use the law of total probability to find the density function of H:

f_H(z) = ∫[0,1] f_H(z | X=x) fX(x) dx

fX(x) = 1 for 0 < x < 1, and fX(x) = 0 otherwise.

f_H(z) = ∫[0,1] 1/(x+2) dx = ln(x+2) |[0,1] = ln(3) - ln(2) = ln(3/2)

Finally, to find P(lnH > z), we integrate the density function of H from z to ∞:

P(lnH > z) = ∫[z,∞] f_H(z) dz = ∫[z,∞] ln(3/2) dz = ln(3/2) * z |[z,∞] = ln(3/2) * ∞ - ln(3/2) * z = ∞ - ln(3/2) * z

Therefore, the probability P(lnH > z) is equal to ∞ - ln(3/2) * z, where z satisfies ez < 2.

Answer 3

[AI-generated answer — Explain Bot]

To find the probability P(ln H ≥ z), where z is a given number satisfying exp(z) < 2, we are asked to use the hint: "Condition on X."

Let's follow the hint and start by conditioning on X. We'll consider the distribution of H given a specific value of X.

For a fixed value of X, H = (X + 2)Y, where X and Y are independent continuous random variables uniformly distributed on (0,1).

To find the probability P(ln H ≥ z), we'll rewrite it as P(H ≥ exp(z)). Now, we can think about this conditional probability in terms of the distribution of Y.

Given a specific value of X, we have H = (X + 2)Y, where X is fixed. To find the distribution of H, we'll need to find the distribution of (X + 2)Y.

Let's first find the cumulative distribution function (CDF) of (X + 2)Y:

F_H(h) = P((X + 2)Y ≤ h) = P(Y ≤ h / (X + 2))

Since Y is uniformly distributed on (0,1), we know that:

P(Y ≤ y) = y if 0 ≤ y ≤ 1, and P(Y ≤ y) = 1 if y > 1.

Therefore, we can write the CDF as:

F_H(h) = { h / (X + 2) if 0 ≤ h / (X + 2) ≤ 1, and 1 if h / (X + 2) > 1.

Now, let's find the probability density function (PDF) of H by taking the derivative of the CDF with respect to h:

f_H(h) = d/dh(F_H(h)) for 0 ≤ h ≤ 1.

Now, we have the PDF of H, but we need to find the PDF of ln(H). To do this, we'll use the change of variables technique.

Let Z = ln(H). Then, H = exp(Z). Taking the derivative with respect to Z, we have:

dH/dZ = d(exp(Z))/dZ = exp(Z).

Now, we can use this change of variables to find the PDF of Z. We'll denote it as f_Z(z):

f_Z(z) = f_H(exp(z)) |d(exp(z))/dz|.
= f_H(exp(z)) |exp(z)|,
= f_H(exp(z)) * exp(z).

Substituting the previously derived f_H(h) into this expression, we have:

f_Z(z) = [f_H(exp(z)) * exp(z)] for 0 ≤ exp(z) ≤ 1.

Now, we have the PDF of Z. However, we are asked to find the PDF fz(z) of Z in terms of the standard normal random variable X, using the fact that Z = F_X(X). Therefore, we need to find the CDF of X, denoted as FX(x).

Given that X is a standard normal random variable, its CDF is already known. Let's denote it as FX(x) = Φ(x), where Φ(x) represents the standard normal CDF.

Now, we can express the PDF fz(z) in terms of the standard normal CDF Φ(x):

fz(z) = [f_H(exp(z)) * exp(z)] for 0 ≤ exp(z) ≤ 1.

However, we want to express this in terms of FX(x) = Φ(x). We'll use the fact that Z = F_X(X) to relate Z and X.

Z = F_X(X)
exp(Z) = F_X(X)
exp(F_X^(-1)(exp(z))) = F_X(X)
F_X^(-1)(exp(z)) = X.

Substituting this into the expression for fz(z), we have:

fz(z) = [f_H(exp(z)) * exp(z)] for 0 ≤ exp(z) ≤ 1
= [f_H(exp(F_X^(-1)(exp(z)))) * exp(F_X^(-1)(exp(z)))] for 0 ≤ exp(z) ≤ 1
= [f_H(F_X^(-1)(exp(z))) * exp(F_X^(-1)(exp(z)))] for 0 ≤ exp(z) ≤ 1
= [f_H(F_X^(-1)(exp(z))) * exp(F_X^(-1)(exp(z)))] for Φ^(-1)(exp(z)) ≤ x ≤ ∞.

Therefore, the PDF fz(z) in terms of the standard normal CDF Φ(x) is given by:

fz(z) = [f_H(F_X^(-1)(exp(z))) * exp(F_X^(-1)(exp(z)))] for Φ^(-1)(exp(z)) ≤ x ≤ ∞.

Please note that we have not provided explicit calculations for f_H(h), as it requires working with the specific distribution of (X + 2)Y given X. However, once you have the distribution of H, you can use the above formula to find the PDF fz(z) in terms of the standard normal CDF Φ(x) by substituting the appropriate values.