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. 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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