Let X and Y be independent continuous random variables that are uniformly distributed on (0,1). Let H=(X+2)Y. Find the probability \mathbf{P}(\ln H\geq z) where z is a given number that satisfies \displaystyle e^ z<2. Your answer should be a function of z.

Hint: Condition on X.

\mathbf{P}(\ln H\geq z)=\quad


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

f_ Z(z)=\quad

1 answer

We can start by finding the probability distribution function (PDF) of H.

Let's first condition on X. Since X is uniformly distributed on (0,1), the conditional PDF of X given a specific value x is 1 for 0 < x < 1 and 0 otherwise.

Next, we can consider the random variable Y. Since Y is also uniformly distributed on (0,1), its PDF is also 1 for 0 < y < 1 and 0 otherwise.

Now, let's find the PDF of H given a specific value x:

Since H = (X+2)Y, we can rewrite this as H = XY + 2Y.

The PDF of XY can be found by convolution:

f_{XY}(h|x) = ∫ f_X(x') f_Y(h/x') dx'

Since X and Y are independent, their joint PDF is simply the product of their individual PDFs:

f_{XY}(h|x) = f_X(x) f_Y(h/x)

Since both X and Y are uniformly distributed on (0,1), their PDFs are both 1 for 0 < x,y < 1. Therefore, the joint PDF simplifies to:

f_{XY}(h|x) = 1

Now, let's consider the random variable 2Y. The PDF of 2Y can be found by a change of variables:

f_{2Y}(h|x) = f_Y(h/2)

Again, since Y is uniformly distributed on (0,1), its PDF is 1 for 0 < y < 1. Therefore, the PDF of 2Y is:

f_{2Y}(h|x) = 1

Finally, to find the PDF of H, we need to convolve the PDFs of XY and 2Y:

f_H(h|x) = ∫ f_{XY}(h-h') f_{2Y}(h') dh'

Since both PDFs are constant (equal to 1), the convolution simplifies to:

f_H(h|x) = area of overlap of [0,1] with [h-1,h] = min(h,1) - max(h-1,0)

In other words, the PDF of H given X = x is:

f_H(h|x) = min(h,1) - max(h-1,0)

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

F_H(h|x) = ∫ f_H(t|x) dt for t from -∞ to h

= ∫ (min(t,1) - max(t-1,0)) dt for t from -∞ to h

= [t^2/2 - (t-1)^2/2] from -∞ to h

= (h^2/2 - (h-1)^2/2) - (0^2/2 - (-1)^2/2)

= (h^2/2 - (h^2 - 2h + 1)/2) - (0 - 1/2)

= (h^2 - h^2 + 2h - 1)/2 + 1/2

= (2h - 1)/2 + 1/2

= h/2 + 1/2

Therefore, the CDF of H given X = x is:

F_H(h|x) = h/2 + 1/2

Now, let's find the CDF of H by conditioning on X:

F_H(h) = ∫ F_H(h|x) f_X(x) dx for x from 0 to 1

= ∫ (h/2 + 1/2) dx for x from 0 to 1

= (hx/2 + x/2) from 0 to 1

= (h/2 + 1/2) - (0/2 + 0/2)

= h/2 + 1/2

Therefore, the CDF of H is:

F_H(h) = h/2 + 1/2

Now, let's find the probability \mathbf{P}(\ln H\geq z):

\mathbf{P}(\ln H\geq z) = \mathbf{P}(H\geq e^z)

Since the CDF of H is F_H(h) = h/2 + 1/2, we can find the probability as:

\mathbf{P}(\ln H\geq z) = 1 - \mathbf{P}(H < e^z)

= 1 - F_H(e^z)

= 1 - (e^z/2 + 1/2)

= 1/2 - e^z/2

Therefore, the probability \mathbf{P}(\ln H\geq z) is 1/2 - e^z/2.