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 f_(Y|X) (y|x)=1/x, 0<=y<=x. Find the PDF f_Y (y) of Y. It will take the form, f_Y (y)={a*log(b) if y belongs to [d,e], a*log(c/y) if y belongs to [e,f], 0 otherwise.

Question

1. Answer by finding a,b,c,d,e,f, where d<e<f. Recall: If 0<=a<b, then integral a to b (1/x) dx = log(b/a).

2. Calculate P(|X-Y|<=1.

3. Calculate P(min(X,Y)<=6

Answer 1

To find the PDF f_Y(y) of Y, we need to consider the different intervals where Y can take on different values.

1. For 0 <= y <= 3:
In this interval, f_Y(y) = 0 since the given conditional PDF only applies when 0 <= y <= x.

2. For 3 <= y <= 8:
In this interval, Y can be any value between Y = 3 and Y = 8. Since X is uniformly distributed on the interval [3,8], the probability that X is less than or equal to x is (x - 3) / (8 - 3) = (x - 3) / 5.
Therefore, for 3 <= y <= 8, f_Y(y) = ∫[y/5 to 1] (1/x) dx = log(5/y).

3. For y >= 8:
In this interval, f_Y(y) = 0 since the given conditional PDF only applies when 0 <= y <= x and X is uniformly distributed on the interval [3,8].

Putting it all together, we have:

f_Y(y) = { log(5/y) if 3 <= y <= 8,
0 otherwise.

To calculate P(|X - Y| <= 1), we need to find the range of x and y values that satisfy this condition. Since X is uniformly distributed on [3,8], X can take on any value between 3 and 8.
For a given value of X = x, Y is uniformly distributed on [0,x], so Y can take on any value between 0 and x.
To satisfy |X - Y| <= 1, we need to find the range of x and y values where either X - Y <= 1 or Y - X <= 1.

X - Y <= 1:
From X - Y <= 1, we get Y >= X - 1. Since Y should be between 0 and x, the range of x and y values that satisfy this condition is when X is between 4 and 8 and Y is between 0 and X - 1.

Y - X <= 1:
From Y - X <= 1, we get Y <= X + 1. Since Y should be between 0 and x, the range of x and y values that satisfy this condition is when X is between 3 and 7 and Y is between 0 and X + 1.

Taking the intersection of these ranges, we find that X should be between 4 and 7 and Y should be between 0 and X - 1.

Therefore, to calculate P(|X - Y| <= 1), we need to find the probability that both X and Y fall within this range.
P(|X - Y| <= 1) = ∫[4 to 7] ∫[0 to x-1] (1/5) dx dy.

To calculate P(min(X,Y) <= 6), we need to find the probability that either X <= 6 or Y <= 6.

P(min(X,Y) <= 6) = P(X <= 6 or Y <= 6) = P(X <= 6) + P(Y <= 6) - P(X <= 6 and Y <= 6).

P(X <= 6) = (6 - 3) / (8 - 3) = 3/5
P(Y <= 6) = ∫[0 to 6] (1/6) dx = 1.

P(X <= 6 and Y <= 6) = ∫[0 to 6] ∫[0 to x] (1/5) dx dy.

Therefore, P(min(X,Y) <= 6) = (3/5) + 1 - ∫[0 to 6] ∫[0 to x] (1/5) dx dy.