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)=1x, 0≤y≤x.
Find the PDF fY(y) of Y. It will take the form
fY(y)=⎧⎩⎨aln(b)aln(c/y)0y∈[d,e]y∈[e,f]otherwise.
Answer by finding a,b,c,d,e,f, where d<e<f.
Recall: If 0≤a<b, then ∫ba1xdx=ln(ba).
(Enter your answers as fractions, or decimals accurate to at least 4 decimal places.)
a= 1/5
b= 8/3
c= 8
d= 0
e= 3
f= 8
Using the given values of a, b, c, d, e, and f:
a = 1/5
b = 8/3
c = 8
d = 0
e = 3
f = 8
We can now rewrite the PDF fY(y) of Y in the given form:
fY(y) = a ln(b) / ln(c/y), for 0 ≤ y ≤ e
fY(y) = 0, otherwise
Plugging in the values:
fY(y) = (1/5) ln(8/3) / ln(8/y), for 0 ≤ y ≤ 3
fY(y) = 0, for y > 3
Therefore, the PDF fY(y) of Y is:
fY(y) = (1/5) ln(8/3) / ln(8/y), for 0 ≤ y ≤ 3
fY(y) = 0, for y > 3
Calculate P(|X−Y|≤1).
To calculate P(|X-Y| ≤ 1), we need to find the probability that the absolute difference between X and Y is less than or equal to 1.
We can start by finding the joint distribution function F(X, Y) of X and Y:
F(X, Y) = P(X ≤ x, Y ≤ y)
Since we already know the conditional PDF of Y given X, we can use it to find the joint CDF:
F(X, Y) = ∫[0, y] ∫[3, x] fY|X(u|x) * fX(v) dv du
F(X, Y) = (1/x) ∫[0, y] ∫[3, x] (1/5) dv du
F(X, Y) = (1/x) * (1/5) * (x-3) * y
F(X, Y) = (1/5) * (x-3) * y/x
Now, we can calculate the probability P(|X-Y| ≤ 1) using the joint CDF.
P(|X-Y| ≤ 1) = P(-1 ≤ X-Y ≤ 1)
This can be rewritten as:
P(X-Y ≤ 1) - P(X-Y ≤ -1)
To find these probabilities, we evaluate the joint CDF at the corresponding values:
P(|X-Y| ≤ 1) = F(X, Y) evaluated at (x, y) = (4, 5) - F(X, Y) evaluated at (x, y) = (4, 3)
P(|X-Y| ≤ 1) = [(1/5) * (4-3) * 5/4] - [(1/5) * (4-3) * 3/4]
P(|X-Y| ≤ 1) = (1/5) * 5 - (1/5) * 3
P(|X-Y| ≤ 1) = 1 - (3/5)
P(|X-Y| ≤ 1) = 2/5
Therefore, the probability P(|X-Y| ≤ 1) is 2/5.
Calculate P(min(X,Y)≤6).
To calculate P(min(X, Y) ≤ 6), we need to find the probability that the minimum value between X and Y is less than or equal to 6.
Since the conditional PDF of Y given X is given as fY|X(y|x) = 1/x, 0 ≤ y ≤ x, we can use this information to calculate the desired probability.
We can rewrite the event "min(X, Y) ≤ 6" as "Y ≤ 6" (because Y will always be less than or equal to X as per the given conditional PDF). Therefore, we need to find P(Y ≤ 6).
We can calculate this probability by integrating the joint PDF fX,Y(x, y) over the region where Y ≤ 6:
P(Y ≤ 6) = ∫[3, 8] ∫[0, 6] fX,Y(x, y) dy dx
P(Y ≤ 6) = ∫[3, 8] ∫[0, 6] (1/(5x)) dy dx
P(Y ≤ 6) = (1/5) ∫[3, 8] ln(x) dy dx
P(Y ≤ 6) = (1/5) ∫[3, 8] 6ln(x) dx
Using integration by parts, we can find the integral:
P(Y ≤ 6) = (1/5) [6xln(x) - 6x]|[3, 8]
P(Y ≤ 6) = (1/5) [(6(8)ln(8) - 6(8)) - (6(3)ln(3) - 6(3))]
P(Y ≤ 6) = (1/5) [48ln(8) - 48 - 18ln(3) + 18]
P(Y ≤ 6) = (1/5) [48ln(8) - 18ln(3) - 30]
P(Y ≤ 6) ≈ 0.335
Therefore, the probability P(min(X, Y) ≤ 6) is approximately 0.335.
To find the PDF fY(y) of Y, we need to determine the value of a, b, c, d, e, and f in the given form.
First, we will find the marginal PDF fX(x) of X.
Since X is uniformly distributed on the interval [3, 8], we know that the PDF of X is:
fX(x) = 1/(8-3) = 1/5, for 3 ≤ x ≤ 8.
Next, we can find the joint PDF fX,Y(x, y) of X and Y using the conditional PDF fY|X(y|x) given in the problem.
fX,Y(x, y) = fY|X(y|x) * fX(x)
fX,Y(x, y) = (1/x) * (1/5) = 1/(5x), for 0 ≤ y ≤ x, 3 ≤ x ≤ 8.
To find the marginal PDF fY(y) of Y, we need to integrate the joint PDF fX,Y(x, y) with respect to x over the range where y is defined:
fY(y) = ∫[3, 8] (1/(5x)) dx, for 0 ≤ y ≤ x
fY(y) = (1/5) ∫[3, 8] (1/x) dx, for 0 ≤ y ≤ x
Using the given recall formula, we can evaluate the integral:
fY(y) = (1/5) ln(x)|[3, 8], for 0 ≤ y ≤ x
fY(y) = (1/5) (ln(8) - ln(3)), for 0 ≤ y ≤ x
fY(y) = (1/5) ln(8/3), for 0 ≤ y ≤ x
Now, let's determine the values of a, b, c, d, e, and f:
a = 1/5
b = ln(8/3)
c = 1
d = 0
e = 0
f = 0 (since the given form indicates that the PDF is 0 for any other value of y)
Therefore, the PDF fY(y) of Y has the form:
fY(y) = (1/5) ln(8/3), for 0 ≤ y ≤ x
This is the final answer.