1. 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 e^z<2. Your answer should be a function of z.
Hint: Condition on X.
P(lnH≥z)=
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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)=
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1. To find the probability P(lnH≥z), we can condition on X. Let's consider the event A = {lnH≥z}. This event occurs if and only if Y ≥ e^(-z/(X+2)).
Since X and Y are independent, we can write the probability as:
P(lnH≥z) = P(A) = ∫[0,1] P(A|X=x) fX(x) dx,
where fX(x) is the probability density function of X.
Now, let's find P(A|X=x):
P(A|X=x) = P(Y ≥ e^(-z/(x+2))) = 1 - P(Y < e^(-z/(x+2))).
Since Y is uniformly distributed on (0,1), the probability P(Y < e^(-z/(x+2))) can be directly calculated as e^(-z/(x+2)).
Therefore, P(A|X=x) = 1 - e^(-z/(x+2)).
Now, let's substitute this back into the integral:
P(lnH≥z) = ∫[0,1] (1 - e^(-z/(x+2))) fX(x) dx.
Since X is uniformly distributed on (0,1), its probability density function is equal to 1 on (0,1) and 0 elsewhere.
Therefore, P(lnH≥z) = ∫[0,1] (1 - e^(-z/(x+2))) dx = ∫[0,1] (1 - e^(-z/(x+2))) dx.
2. Let's find the PDF fZ(z) of the random variable Z = FX(X), where X is a standard normal random variable.
The cumulative distribution function (CDF) of Z can be written as:
FZ(z) = P(Z ≤ z) = P(FX(X) ≤ z).
Now, let's use the inverse of the CDF FX(x) to find the connection between X and Z:
FX(X) = z is equivalent to X = FX^(-1)(z).
Therefore, the probability density function (PDF) of Z can be obtained as:
fZ(z) = d/dz P(Z ≤ z) = d/dz P(FX(X) ≤ z) = d/dz P(X ≤ FX^(-1)(z)).
Since X is a standard normal random variable, its PDF fX(x) is given by:
fX(x) = (1/√(2π)) * e^(-x^2/2).
Now, let's use the chain rule to differentiate the CDF:
fZ(z) = d/dz P(X ≤ FX^(-1)(z)) = d/dz ∫[-∞, FX^(-1)(z)] fX(x) dx.
Let's denote FX^(-1)(z) as x0. Then, we can rewrite the integral as:
fZ(z) = d/dz ∫[-∞, x0] fX(x) dx.
Since the limits of integration are constant, we can differentiate under the integral sign:
fZ(z) = ∫[-∞, x0] d/dz fX(x) dx.
Now, let's differentiate the normal distribution PDF fX(x) with respect to z:
d/dz fX(x) = d/dz ((1/√(2π)) * e^(-x^2/2)) = (-x/√(2π)) * e^(-x^2/2).
Substituting this back into the integral, we have:
fZ(z) = ∫[-∞, x0] (-x/√(2π)) * e^(-x^2/2) dx.
Therefore, the PDF fZ(z) of the random variable Z = FX(X) is (-x/√(2π)) * e^(-x^2/2) integrated over the appropriate interval [-∞, x0].
1. To find the probability P(lnH≥z), let's condition on the value of X.
Since X and Y are independent, we can write the joint density function as f(x, y) = fX(x)fY(y) where fX(x) and fY(y) are the probability density functions of X and Y, respectively.
Given that X and Y are uniformly distributed on (0,1), their probability density functions are both equal to 1 over the interval (0,1).
So, fX(x) = 1 for 0 < x < 1 and fY(y) = 1 for 0 < y < 1.
Now, let's find the cumulative distribution function (CDF) of H.
F(z) = P(lnH ≤ z) = P((X + 2)Y ≤ e^z)
Let's condition on X.
F(z) = ∫[0,1] P((X + 2)Y ≤ e^z | X = x) * fX(x) dx
Since Y is also uniformly distributed on (0,1), we can write the conditional probability as:
P((X + 2)Y ≤ e^z | X = x) = P(Y ≤ e^z / (x + 2))
Since Y is uniformly distributed on (0,1), P(Y ≤ e^z / (x + 2)) equals e^z / (x + 2) for 0 < e^z / (x + 2) < 1, and 1 for e^z / (x + 2) ≥ 1.
Therefore, the conditional probability density function of H given X = x is:
fH|X(h | x) = e^z / (x + 2) for 0 < h < e^z / (x + 2)
1 for h ≥ e^z / (x + 2)
To find the CDF of H, integrate over the conditional probability density function:
F(z) = ∫[0,e^z] fH|X(h | x) dh + ∫[e^z,∞] fH|X(h | x) dh
Integrating the first term:
∫[0,e^z] e^z / (x + 2) dh = e^z / (x + 2) * h |[0,e^z]
= e^z / (x + 2) * e^z
= e^(2z) / (x + 2)
Integrating the second term:
∫[e^z,∞] 1 dh = h |[e^z,∞]
= ∞ - e^z
= ∞ (since e^z < ∞)
Therefore, the cumulative distribution function (CDF) is:
F(z) = e^(2z) / (x + 2) for 0 < z < ln(2)
1 for z ≥ ln(2)
To find the probability P(lnH ≥ z), we need to find 1 - F(z):
P(lnH ≥ z) = 1 - F(z)
Therefore, the probability P(lnH ≥ z) is:
P(lnH ≥ z) = 1 - e^(2z) / (x + 2) for 0 < z < ln(2)
0 for z ≥ ln(2)
2. The random variable Z = FX(X) is the CDF of the standard normal random variable X.
To find the probability density function (PDF) of Z, we differentiate the CDF with respect to z:
fZ(z) = d/dz FX(X)
The PDF of the standard normal distribution is φ(x) = (1/√(2π)) * e^(-x^2/2).
So, to find fZ(z), we substitute z with FX^(-1)(z) in the PDF of the standard normal distribution:
fZ(z) = φ(FX^(-1)(z))
= (1/√(2π)) * e^(-[FX^(-1)(z)]^2/2)
Therefore, the PDF fZ(z) of Z is:
fZ(z) = (1/√(2π)) * e^(-[FX^(-1)(z)]^2/2)
Note: The PDF fZ(z) takes values between 0 and 1.