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.
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
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.
To find the PDF f_Y(y) of Y, we need to consider the different intervals where the probability density function is defined.
1. Calculation of a, b, c, d, e, f:
Since Y is uniformly distributed on the interval [0,x] given X=x, the probability density function for Y given X=x is f_(Y|X)(y|x) = 1/x for 0 <= y <= x.
To find the PDF f_Y(y) of Y, we need to calculate the probability density function for Y irrespective of the value of X.
Let's consider the intervals:
Interval [0, d):
Since Y cannot take values less than 0, the probability density function for this interval is 0.
Interval [d, e]:
For this interval, we need to integrate the conditional PDF f_(Y|X)(y|x) = 1/x over the range x = d to x = e. Since f_(Y|X)(y|x) = 1/x, the integral is given by:
∫[d, e] (1/x) dx = log(e/d).
Therefore, the PDF for this interval is a * log(b), where a = 1 and b = e/d.
Interval [e, f]:
For this interval, we again need to integrate the conditional PDF f_(Y|X)(y|x) = 1/x over the range x = e to x = f:
∫[e, f] (1/x) dx = log(f/e).
Therefore, the PDF for this interval is a * log(c/y), where a = 1 and c = f/e.
Putting it all together, we have:
f_Y(y) = { a * log(b) if y belongs to [d, e]
{ a * log(c/y) if y belongs to [e, f]
{ 0 otherwise.
2. Calculation of P(|X-Y| <= 1):
To calculate this probability, we need to find the region where |X-Y|<=1.
|X-Y|<=1 means that X-Y<=1 and Y-X<=1. Rearranging the inequalities, we get:
-1<=X-Y<=1 and -1<=Y-X<=1.
Adding these two conditions, we have -2<=X-Y+Y-X<=2, which simplifies to -2<=0<=2.
Therefore, the probability P(|X-Y|<=1) is 1, as the probability of an event that is guaranteed to occur is 1.
3. Calculation of P(min(X,Y) <= 6):
To calculate this probability, we need to find the region where min(X,Y)<=6.
Since min(X,Y) is the smaller of the two numbers X and Y, we need to consider two cases:
Case 1: min(X,Y) = X.
In this case, X <= 6. The probability of this event is the probability of X being less than or equal to 6, which is (6-3)/(8-3) = 3/5.
Case 2: min(X,Y) = Y.
In this case, Y <= 6. The probability of this event is the probability of Y being less than or equal to 6, which is (6-0)/(8-3) = 6/5.
Since min(X,Y) can be either X or Y, we need to consider both cases. As these events are mutually exclusive, we can add their probabilities:
P(min(X,Y) <= 6) = P(X <= 6 or Y <= 6) = P(X <= 6) + P(Y <= 6) = 3/5 + 6/5 = 9/5.