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)=
2. Let X be a standard normal random variable, and let F(x) be its CDF, consider the random variable Z=F(X). Find the PDF f(z) of Z. Note that f(z)
takes values in (0,1).
f(z)=
1. P(lnH ≥ z) = P(H ≥ e^z)
Since H = max(X+2, Y), let's consider the events A = {X+2 ≥ e^z} and B = {Y ≥ e^z}. Since X and Y are independent, we can use the fact that P(A ∩ B) = P(A)P(B) for independent events.
P(H ≥ e^z) = P(A ∩ B) = P(A)P(B)
Now, let's find P(A) and P(B).
P(A) = P(X+2 ≥ e^z) = P(X ≥ e^z - 2)
Since X is uniformly distributed on (0,1), the probability of X being greater than or equal to a given value is equal to 1 minus that value.
P(A) = 1 - (e^z - 2) = 3 - e^z
P(B) = P(Y ≥ e^z)
Similarly, Y is also uniformly distributed on (0,1), so
P(B) = 1 - e^z
Therefore, P(lnH ≥ z) = P(A)P(B) = (3 - e^z)(1 - e^z)
2. To find the PDF f(z) of Z, we can use the fact that Z = F(X), where F(x) is the CDF of X. The CDF of a standard normal random variable X is given by F(x) = Φ(x), where Φ(x) is the standard normal cumulative distribution function.
To find the PDF of Z, we need to differentiate the cumulative distribution function F(z) with respect to z:
f(z) = d/dz [F(z)]
Since F(z) = Φ(z), we have:
f(z) = d/dz [Φ(z)]
The derivative of the standard normal cumulative distribution function Φ(z) is the standard normal probability density function φ(z), so:
f(z) = φ(z)
Therefore, the PDF f(z) of Z is equal to the standard normal probability density function φ(z).
1. To find the probability P(lnH ≥ z), we need to consider the probability of the event lnH ≥ z given the value of X. Since X and Y are independent continuous uniform random variables on (0, 1), we can write the joint PDF (probability density function) as:
fXY(x, y) = fX(x) * fY(y) = 1 * 1 = 1, for 0 < x < 1 and 0 < y < 1.
To find the conditional PDF of Y given X = x, we need to determine the joint PDF of Y and lnH and then integrate out X.
Joint PDF of lnH and Y given X = x:
f(lnH, y | X = x) = fXY(x, y) / fX(x)
= 1 / 1
= 1, for 0 < y < 1.
Now, we can find the probability P(lnH ≥ z) by integrating the conditional PDF over the appropriate range:
P(lnH ≥ z) = ∫∫[z, 1]f(lnH, y | X = x)dydx
= ∫_[0,1] ∫_[z,1] 1 dy dx
= ∫_[0,1] (1 - z) dx
= (1 - z) ∫_[0,1] 1 dx
= (1 - z) [x]_|[0,1]
= 1 - z.
Therefore, the probability P(lnH ≥ z) is equal to 1 - z, as long as e^z < 2 (which ensures that lnH is a valid quantity).
2. To find the PDF f(z) of the random variable Z = F(X), where X is a standard normal random variable, we need to calculate the derivative of the cumulative distribution function (CDF) F(x) with respect to z.
Since X is a standard normal random variable, its CDF is given by:
F(x) = ∫_-∞^x fX(t) dt = Φ(x), where Φ(x) is the standard normal CDF.
Now, we can express Z = F(X) as:
Z = F(X) = Φ(X).
To find the PDF of Z, we can differentiate the CDF with respect to z:
f(z) = d/dz [F(x)]
= d/dz [Φ(x)].
Unfortunately, there is no closed-form expression for the derivative of the standard normal CDF. However, you can approximate it using numerical methods or use standard normal distribution tables to estimate the value of f(z) for specific z-values.
Note that f(z) takes values in (0, 1) since it represents a probability density, and the area under the curve integrated over its entire range should equal 1.
1. To find the probability P(lnH ≥ z), we can condition on the random variable X.
Since X and Y are independent continuous random variables uniformly distributed on (0,1), we can express the joint probability density function (pdf) as:
fXY(x, y) = 1, for 0 < x < 1 and 0 < y < 1
Now let's find the cumulative distribution function (cdf) of H:
F(H ≤ h) = P(H ≤ h) = P((X + 2)Y ≤ h)
= P(Y ≤ h / (X + 2))
= ∫[0, 1] P(Y ≤ h / (x + 2)) * fX(x) dx
= ∫[0, 1] ∫[0, h / (x + 2)] 1 * dy * fX(x) dx
= ∫[0, 1] (h / (x + 2)) * fX(x) dx
Since X is uniformly distributed on (0,1), its pdf is given by fX(x) = 1, for 0 < x < 1.
Therefore, the cdf of H is:
F(H ≤ h) = ∫[0, 1] (h / (x + 2)) dx
= h * ∫[0, 1] 1 / (x + 2) dx
= h * ln(x + 2) ∣[0, 1]
= h * ln(3)
To find the pdf of H, we differentiate the cdf with respect to h:
fH(h) = d/dh [F(H ≤ h)]
= d/dh [h * ln(3)]
= ln(3)
Now let's find the probability P(lnH ≥ z):
P(lnH ≥ z) = 1 - P(lnH < z)
= 1 - F(lnH ≤ z)
= 1 - F(H ≤ e^z)
= 1 - e^z * ln(3)
Therefore, the probability P(lnH ≥ z) is given by:
P(lnH ≥ z) = 1 - e^z * ln(3)
2. To find the probability density function (pdf) f(z) of Z = F(X), we can use the transformation method for random variables.
The standard normal random variable X has a cumulative distribution function (CDF) F(x) = Φ(x), where Φ(x) is the standard normal CDF.
Now let's find the CDF of Z = F(X):
FZ(z) = P(Z ≤ z) = P(F(X) ≤ z)
= P(X ≤ F^(-1)(z))
= Φ(F^(-1)(z))
To find the pdf f(z) = d/dz [FZ(z)], we differentiate the CDF with respect to z:
f(z) = d/dz [Φ(F^(-1)(z))]
However, there isn't a closed-form expression for the derivative of the standard normal CDF Φ(x). Therefore, there isn't a simple closed-form expression for the pdf f(z) of Z = F(X).
Instead, we can use numerical methods or simulation techniques to approximate the pdf f(z) of Z.