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.
P(lnH≥z1)= ?
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)= ?
Please, can someone share me the correct answers? I've spent a lot of time trying to solve this problem, but I can't.
if X and Y are uniform random (0,1) and H=(X+2)Y, then the distribution for H looks like the piecewise graph y={0.4: 0<x<2, 1.2-.4x, 2<x<3}, which i estimated using simulation
upon further consideration (using math not simulation), the function for H is
{0<x<2: ln(3)-ln(2), 2<x<3: ln(3)-ln(x)} (piecewise)
the graph of H:
desmos.com/calculator/jf384j0mrt
the probability graph that P(ln(H)>z):
desmos.com/calculator/io40jln2nx
1. P(lnH≥z1)= 1-e^z*ln(3/2)
2. fZ(z)= 1
Why don't mathematicians bring a ladder to the bar? Because everyone always has a lot of problems to tackle, but they never seem to reach a solution!
Now, let's tackle these probability problems together, shall we?
For the first problem, let's condition on X. We know that H=(X+2)Y. So, we can rewrite the probability as:
P(lnH≥z) = P(ln((X+2)Y)≥z)
Since X and Y are independent, we can split the probability:
P(ln((X+2)Y)≥z) = P(ln(X+2)+ln(Y)≥z)
Now, let's consider the cumulative distribution function (CDF) of X and Y. Since they are uniformly distributed on (0,1), their CDFs are simply:
FX(x) = x for 0 < x < 1
FY(y) = y for 0 < y < 1
Now, we apply the natural logarithm to both sides of the inequality:
ln(X+2)+ln(Y)≥z
Taking the exponential of both sides, we get:
(X+2)Y≥e^z
Notice that e^z is less than 2 by the given condition, ez < 2. So, we have:
P(lnH≥z) = P((X+2)Y≥e^z)
To solve this probability, we need to integrate over the joint probability density function (PDF) of X and Y:
P(lnH≥z) = ∫∫(X+2)Y * fXY(x,y) dxdy
However, since X and Y are independent, the joint PDF simplifies:
P(lnH≥z) = ∫∫(X+2)Y * fX(x) * fY(y) dxdy
Since fX(x) = 1 for 0 < x < 1, and fY(y) = 1 for 0 < y < 1, we have:
P(lnH≥z) = ∫∫(X+2)Y dxdy
Now, you can calculate this integral to find the probability P(lnH≥z).
As for the second problem, let's find the PDF fZ(z) of the random variable Z=FX(X).
To do this, we need to find the cumulative distribution function (CDF) of Z:
FZ(z) = P(Z≤z) = P(FX(X)≤z) = P(X≤FX^(-1)(z))
Since X is a standard normal random variable, its inverse CDF is simply the standard normal distribution function:
FX^(-1)(z) = Φ^(-1)(z)
So, we have:
FZ(z) = P(X≤Φ^(-1)(z)) = Φ(Φ^(-1)(z))
Taking the derivative of FZ(z) with respect to z, we get the PDF fZ(z):
fZ(z) = d/dz[Φ(Φ^(-1)(z))]
I hope these explanations help you solve the problems. Don't Clown Bot worry, you'll get it!
To find the probability P(lnH≥z) where z is a given number, we can start by conditioning on X.
Since X and Y are independent and uniformly distributed on (0,1), X follows a continuous uniform distribution on (0,1) with probability density function (PDF) fX(x) = 1 for 0 < x < 1.
Let's find the cumulative distribution function (CDF) and the probability density function (PDF) of H.
We have H = (X+2)Y.
To find the CDF of H, we need to condition on X:
F_H(z) = P(H ≤ z)
= P((X+2)Y ≤ z | X)
= ∫[0,1] P((X+2)Y ≤ z | X=x) * fX(x) dx
= ∫[0,1] P(Y ≤ z/(x+2)) * fX(x) dx
= ∫[0,1] (z/(x+2)) * fX(x) dx
= ∫[0,1] (z/(x+2)) dx
= [z * ln(x+2)] [0,1]
= z * ln(3) - z * ln(2)
Now, P(lnH≥z) = 1 - P(lnH < z) = 1 - F_H(e^z) = 1 - (e^z * ln(3) - e^z * ln(2)).
So, the probability P(lnH≥z) is given by 1 - (e^z * ln(3) - e^z * ln(2)) where ez < 2.
Now let's move on to the second question.
Let X be a standard normal random variable with CDF FX(x).
We define the random variable Z = FX(X).
To find the PDF fZ(z) of Z, we can use the probability integral transform.
The probability integral transform states that if X is a continuous random variable with CDF FX(x), then the random variable Y = FX(X) has a uniform distribution on (0,1).
Since X is a standard normal random variable, FX(x) is the standard normal CDF, which is a well-known function.
Therefore, we have Z = FX(X) ~ Uniform(0,1).
The uniform distribution on (0,1) has a constant PDF of 1 for 0 < z < 1.
So, the PDF fZ(z) of Z is fZ(z) = 1 for 0 < z < 1.
I hope this explanation helps you understand how to solve the problems. If you have any further questions or need clarification, please let me know!