Suppose that X is uniformly distributed on the interval [3,8] , and that given X=x , Y is uniformly distributed on the interval [0,x] . That is, the conditional PDF of Y given X=x is

Question

fY|X(y|x)=1/x, 0≤y≤x.

Find the PDF fY(y) of Y . It will take the form

fY(y)=⎧⎩⎨aln(b) y∈[d,e]
aln(c/y) y∈[e,f]
0 otherwise.
Answer by finding a,b,c,d,e,f , where d<e<f .

Recall: If 0≤a<b, then ∫ba 1/xdx=ln(ba) .

Continue from the problem above, i.e. X and Y are defined as above.

Calculate P(|X−Y|≤1).
Calculate P(min(X,Y)≤6).

Answer 1

To find the PDF of Y, we need to consider the range of possible values for Y and determine the corresponding probability density function.

Given that X is uniformly distributed on the interval [3, 8], the probability density function of X is:
fX(x) = 1/(8-3) = 1/5, for 3 ≤ x ≤ 8

We will consider the cases where Y falls within different intervals with respect to X.

Case 1: 0 ≤ y ≤ 3
In this case, the probability that Y takes a value between 0 and 3 is 1/5, regardless of the value of X. Therefore, fY(y) = 1/5 for 0 ≤ y ≤ 3.

Case 2: 3 < y ≤ 8
In this case, the probability that Y takes a value between 3 and 8 is given by the conditional PDF fY|X(y|x) = 1/x. Since Y is uniformly distributed on the interval [0, X] given X = x, the cumulative distribution function (CDF) of Y is given by integrating the conditional PDF over the range [0, y] with respect to X, and then taking the derivative with respect to y.

F(Y ≤ y) = ∫[0,y] fY|X(t|x) dt
= ∫[0,y] 1/x dt
= ln(x)|[0,y]
= ln(y)

Taking the derivative, we obtain:

fY(y) = d/dy (ln(y))
= 1/y

Therefore, the probability density function of Y is fY(y) = 1/y for 3 < y ≤ 8.

Combining the two cases, we can write the PDF of Y as:

fY(y) = ⎧⎩⎨1/5 0 ≤ y ≤ 3,
⎧⎩⎨1/y 3 < y ≤ 8,
0 otherwise.

To calculate P(|X - Y| ≤ 1), we need to consider the joint distribution of X and Y. The probability can be obtained by integrating the joint PDF over the region defined by |x - y| ≤ 1.
P(|X - Y| ≤ 1) = ∫∫ |x - y| ≤ 1 fX(x)fY(y) dx dy

Integrating over the region defined by |x - y| ≤ 1, we split the integral into two parts: x < y + 1 and y - 1 < x.

P(|X - Y| ≤ 1) = ∫∫ x < y + 1 fX(x)fY(y) dx dy + ∫∫ y - 1 < x fX(x)fY(y) dx dy

Within the first integral, the limits of integration for x are 3 to y + 1, and for y they are 0 to 8.
For the second integral, the limits of integration for x are y - 1 to 8, and for y they are 0 to 8.

P(|X - Y| ≤ 1) = ∫[0,8] ∫[3,y+1] fX(x)fY(y) dx dy + ∫[0,8] ∫[y-1,8] fX(x)fY(y) dx dy

Plugging in the expressions for fX(x) and fY(y) from before, we get:

P(|X - Y| ≤ 1) = ∫[0,8] ∫[3,y+1] (1/5)(1/y) dx dy + ∫[0,8] ∫[y-1,8] (1/5)(1/y) dx dy

Evaluating the integrals, we find:

P(|X - Y| ≤ 1) = ln(9/8) + ln(8/7) + ln(7/6) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1)

To calculate P(min(X,Y) ≤ 6), we need to consider the joint distribution of X and Y again. The probability can be obtained by integrating the joint PDF over the region defined by min(x, y) ≤ 6.
P(min(X, Y) ≤ 6) = ∫∫ min(x, y) ≤ 6 fX(x)fY(y) dx dy

Integrating over the region defined by min(x, y) ≤ 6, we split the integral into two parts: x ≤ 6 and y ≤ 6. This gives:

P(min(X, Y) ≤ 6) = ∫[0,6] ∫[0,y] fX(x)fY(y) dx dy + ∫[6,8] ∫[0,6] fX(x)fY(y) dx dy

Using the expressions for fX(x) and fY(y) from before, we have:

P(min(X, Y) ≤ 6) = ∫[0,6] ∫[0,y] (1/5)(1/y) dx dy + ∫[6,8] ∫[0,6] (1/5)(1/y) dx dy

Evaluating the integrals, we find:

P(min(X, Y) ≤ 6) = ln(7/2) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1)

Therefore, P(|X - Y| ≤ 1) = ln(9/8) + ln(8/7) + ln(7/6) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1) and P(min(X, Y) ≤ 6) = ln(7/2) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1).