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
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) .
a=?
b=?
c=?
d=?
e=?
f=?
a=1/5
b=8
c=8
d=0
e=3
f=8
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
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).
Any advice on this?
Calculate P(|X−Y|≤1).
Calculate P(min(X,Y)≤6).
To find the PDF fY(y) of Y, we need to consider the range of y values and the conditional PDF given X=x.
Given that X is uniformly distributed on the interval [3,8], the probability density function (PDF) of X is given by fX(x) = 1/(8-3) = 1/5 for 3 ≤ x ≤ 8.
Now, let's find the range of y values for each interval.
For 0 ≤ y ≤ 3, the PDF fY(y) is 0 because the conditional PDF given X=x is 0 for y less than 0 or greater than x. So, d = 0.
For 3 ≤ y ≤ x, we have the condition that 0 ≤ y ≤ x. Since X is uniformly distributed on [3,8], the lower bound of x is 3. Therefore, e = 3.
For y > x, the conditional PDF fY|X(y|x) is also 0. So, f = 8 (the upper bound of x).
Now, let's find the PDF of Y for the intervals [d,e] and [e,f].
For y ∈ [d,e] = [0,3]:
∫ fY(y) dy = ∫ aln(b) dy from 0 to 3
To find a and b, we need to integrate this PDF over the interval [0, 3]. We know that the integral of the PDF over the interval should be equal to 1, since it represents the probability.
∫₀³ aln(b) dy = 1
Using the integral formula ∫ a ln(b) dy = y ln(b) - y | [0, 3], we get:
3a ln(b) - 3 ln(b) = 1
Simplifying this equation, we have:
ln(b^3) = 1 + 3 ln(b)
Applying exponential function to both sides, we get:
b^3 = e * e^(3 ln(b))
b^3 = e * b^3
We can conclude that e = 1.
For y ∈ [e,f] = [1,8]:
∫ fY(y) dy = ∫ aln(c/y) dy from 1 to 8
Using the integral formula ∫ a ln(c/y) dy = - a ln(y) + y ln(c) | [1, 8], we get:
-a ln(8) + 8 ln(c) + a(ln(1) - ln(c))
Simplifying, we get:
-a ln(8) + 8 ln(c) - a ln(c)
Setting this equal to 1, we have:
-a ln(8) + 8 ln(c) - a ln(c) = 1
Rearranging, we get:
8 ln(c) - a ln(8) - a ln(c) = 1
By comparing coefficients, we can deduce:
8 ln(c) = 1
-ln(8) = -a
Solving this equation, we find:
c = e^(1/8)
a = ln(8)
Therefore, the values are:
a = ln(8)
b = e^(1/3)
c = e^(1/8)
d = 0
e = 3
f = 8
So, the PDF fY(y) of Y takes the form:
fY(y) = ln(8) when y ∈ [0,3]
ln(e^(1/8)/y) when y ∈ [3,8]
0 otherwise.