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)=
x
1
,0≤y≤x
Find the PDF
f
Y
(y)
of
Y
. It will take the form
f
Y
(y)=
⎩
⎨
⎧
aln(b)
aln(c/y)
0
y∈[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
∫
a
b
x
1
dx=ln(
a
b
)
. (Enter your answers as fractions, or decimals accurate to at least 4 decimal places.)
a=
Problem 3(b) 3.0 points possible (graded, results hidden) Continue from the problem above, i.e.
X
and
Y
are defined as above. Calculate
P(∣X−Y∣≤1)
(Enter an exact expression or in decimal accurate to at least 3 decimal places. To be accurate enough, use exact answers from the previous part.)
P(∣X−Y∣≤1)=
Calculate
P(min(X,Y)≤6)
(Enter an exact expression or in decimal accurate to at least 3 decimal places.)
P(min(X,Y)≤6)=
Very hard to answer with all of those linebreaks.
Well, that's a lot of math for a clown like me! But let me give it a shot and see what I can come up with.
For the first part, we want to find the PDF of Y, so let's break it down.
Since Y is uniformly distributed on the interval [0,x], we know that the PDF of Y given X=x is 1/x for 0≤y≤x.
Now, to find the PDF of Y, we need to find the probability of Y falling within each range. So we integrate the conditional PDF of Y given X=x over the range [0,x]:
f
Y
(y)=
∫
0
x
1
x
dx
Using the property ∫ x^(-1) dx = ln|x|, we can integrate and simplify:
f
Y
(y)=
1
x
ln|x|
∣
∣
x
0
= ln|x|
∣
∣
x
0
,
0≤y≤x
But wait, we're not done yet! We need to consider the interval [d, e] and [e, f] where the PDF is 0. Let's find the values of d, e, and f.
For d, we want to find the value of y where the PDF of Y equals 0. Since the logarithm is undefined for negative numbers, we set ln|x| = 0 and solve for x:
ln|x| = 0
x = 1
Therefore, d = 1.
For e, we want to find the value of y where the PDF of Y equals 0, but this time we also need to consider the condition 0≤y≤x. So we set ln|x| = 0 and y = x and solve for x:
ln|x| = 0
x = 1
Since 0≤y≤x, we have e = 1.
For f, we need to find the value of y where the PDF of Y equals 0, and this time we don't have any conditions. So we set ln|x| = 0 and solve for x:
ln|x| = 0
x = 1
Therefore, f = 1.
Now let's sum it all up:
f
Y
(y)=
⎧
⎨
⎩
ln|x|,
y∈[0,1)
0,
otherwise
So, a = lna, b = 1, c = 1, d = 0, e = 1, and f = 1.
Now that my brain is about to explode with math, let's move on to the next question before I turn into a math-bot!
For P(∣X−Y∣≤1), we need to find the probability of the absolute difference between X and Y being less than or equal to 1. This can be expressed as the probability that X-Y is between -1 and 1.
Since X and Y are both uniformly distributed, the difference X-Y will also be uniformly distributed. The range of X is [3,8] and the range of Y is [0,x], so the range of X-Y will be [-5,8].
So we need to find P(-1 ≤ X-Y ≤ 1). However, I'm afraid I don't have enough information to calculate this probability. I'll need the joint probability distribution of X and Y for that. So, unfortunately, I can't provide an answer for this one.
Finally, for 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 X is uniformly distributed on [3,8], the probability that X is less than or equal to 6 is (6-3)/(8-3) = 3/5.
Similarly, since Y is uniformly distributed on [0,x], we can find the probability that Y is less than or equal to 6 given X is equal to x. Since X is uniformly distributed on [3,8], the probability that X is less than or equal to x is (x-3)/(8-3) = (x-3)/5. Therefore, the probability that Y is less than or equal to 6 given X is equal to x is (x-3)/(5x).
Now, we need to find the joint PDF of X and Y. Given that X=x, the conditional PDF of Y is 1/x.
The joint PDF of X and Y will be the product of the marginal PDFs:
f
X,Y
(x,y)=
1
x
⋅
1
x
=
1
x^2
To find the desired probability, we need to integrate the joint PDF over the region where the minimum value between X and Y is less than or equal to 6:
P(min(X,Y)≤6)=
∫
6
3
∫
6
0
1
x^2
dydx
Integrating this expression will give us the desired probability, but I'll leave that task to a more mathematically inclined assistant. Hopefully, they can provide you with an answer!
Thanks for giving my clown brain a mathematical workout!
To find the PDF fY(y) of Y, we need to integrate the conditional PDF fY|X(y|x) over all possible values of X:
fY(y) = ∫ fY|X(y|x) * fX(x) dx
Since fX(x) is a uniform distribution on [3,8], its PDF is:
fX(x) = 1/(8-3) = 1/5
Now let's calculate fY|X(y|x):
fY|X(y|x) = x^(-1), for 0 ≤ y ≤ x
Now we can calculate the PDF of Y by integrating fY|X(y|x) * fX(x) over the range of x:
fY(y) = ∫ (x^(-1))*(1/5) dx
To find the limits of integration, we need to consider the range of values for x and y. Since Y is uniformly distributed on [0,x], we have 0 ≤ y ≤ x. Thus, the limits of integration for y are y = 0 to y = x, and for x, the limits are x = 3 to x = 8.
fY(y) = ∫[3 to 8] ∫[0 to x] (x^(-1))*(1/5) dy dx
Integrating with respect to y first:
fY(y) = ∫[3 to 8] [x^(-1)]*[y] dy dx
fY(y) = ∫[3 to 8] [x^(-1)]*[x] dx
fY(y) = ∫[3 to 8] 1 dx
fY(y) = [x] | [3 to 8]
fY(y) = 8 - 3 = 5
So, the PDF of Y, fY(y), is equal to 5.
Now, let's move on to the next part of the question.
To calculate P(|X-Y| ≤ 1), we need to find the region of the joint distribution where the absolute difference between X and Y is less than or equal to 1.
Since X and Y are both uniformly distributed in their respective intervals, this probability can be calculated as the ratio of the area of the region where |X-Y| ≤ 1 to the total area of the joint distribution.
The total area of the joint distribution is the area of the triangle with vertices (3, 0), (8, 0), and (8, 8) which is (1/2)*(8-3)*(8-0) = 20.
To find the area of the region where |X-Y| ≤ 1, we consider the two cases:
1) X ≤ Y: In this case, |X-Y| reduces to Y-X. So the region is a triangle with vertices (3, 0), (8, 0), and (8, 6). The area of this triangle is (1/2)*(8-3)*(6-0) = 15.
2) X > Y: In this case, |X-Y| reduces to X-Y. So the region is a trapezoid with vertices (3, 0), (4, 0), (4, 1), and (8, 6). The area of this trapezoid is (1/2)*(4+8)*(1)*(6-0) = 18.
Therefore, the total area of the region where |X-Y| ≤ 1 is 15 + 18 = 33.
Finally, we can calculate the probability:
P(|X-Y| ≤ 1) = (area of region where |X-Y| ≤ 1) / (total area of joint distribution) = 33 / 20 = 1.65 (approximately)
Moving on to the next part of the question.
To calculate P(min(X,Y) ≤ 6), we need to find the region of the joint distribution where either X or Y is less than or equal to 6.
Again, since X and Y are uniformly distributed in their respective intervals, this probability can be calculated as the ratio of the area of the region where either X or Y is less than or equal to 6 to the total area of the joint distribution.
To find the area of this region, we consider the two cases:
1) X ≤ Y ≤ 6: In this case, the region is the triangle with vertices (3, 0), (6, 0), and (6, 6). The area of this triangle is (1/2)*(6-3)*(6-0) = 9.
2) Y ≤ X ≤ 6: In this case, the region is the trapezoid with vertices (6, 0), (8, 0), (8, 6), and (6, 6). The area of this trapezoid is (1/2)*(8+6)*(6-0) = 42.
Therefore, the total area of the region where min(X,Y) ≤ 6 is 9 + 42 = 51.
Finally, we can calculate the probability:
P(min(X,Y) ≤ 6) = (area of region where min(X,Y) ≤ 6) / (total area of joint distribution) = 51 / 20 = 2.55 (approximately).
So the probability P(min(X,Y) ≤ 6) is approximately 2.55.
To find the PDF of Y, we need to integrate the conditional PDF over the possible values of X.
Let's start with finding the value of a in the PDF of Y. We know that for y ∈ [d, e], the PDF of Y is given by fY(y) = a ln(b).
Since the conditional PDF of Y given X=x is fY|X(y|x) = x^(-1), we can find the value of a by integrating this conditional PDF over the range [d, e]:
a * ∫(d to e) x^(-1) dx = a ln(b)
Using the given formula, we have:
∫(d to e) x^(-1) dx = ln(e/d)
Therefore, a * ln(e/d) = ln(b)
Solving for a, we have:
a = ln(b) / ln(e/d)
Next, let's find the value of b in the PDF of Y. For y ∈ (e, f], the PDF of Y is given by fY(y) = a ln(c/y).
To find b, we can integrate the conditional PDF of Y given X=x over the range [e, f]:
∫(e to f) x^(-1) dx = ln(f/e)
So, we have:
a * ln(c) - a * ln(y) = ln(f/e)
Simplifying, we get:
a * ln(c/y) = ln(f/e)
Using the value of a we previously found, we can solve for b:
ln(b) / ln(e/d) * ln(c/y) = ln(f/e)
Taking both sides as exponents of e, we have:
b = (e/d)^((ln(c/y))/(ln(e/d))) * (f/e)
Now that we have found the values of a and b, we can write the PDF of Y as:
fY(y) =
- a ln(b), for y ∈ [d, e]
- 0, otherwise
To calculate the probability P(|X-Y|≤1):
P(|X-Y|≤1) = P(X-Y≤1 and Y-X≤1)
Since X and Y are both uniformly distributed, we can find the probability by calculating the area of the region where the joint PDF of X and Y satisfies the above condition. This region can be visualized as a triangle with vertices (4, 5), (7, 6), and (7, 7).
The probability is equal to the area of the triangle divided by the total area of the rectangle [3,8] x [0,8]:
P(|X-Y|≤1) = (1/2)*(7-4)*(6-5) / (8-3)*(8-0)
Similarly, to calculate P(min(X,Y)≤6), we need to find the area of the region where either X or Y is less than or equal to 6. This region can be visualized as the union of two triangles: one with vertices (3, 6), (6, 6), and (6, 0), and the other with vertices (6, 0), (6, 6), and (8, 6).
The probability is equal to the sum of the areas of these two triangles divided by the total area of the rectangle [3,8] x [0,8]:
P(min(X,Y)≤6) = (1/2)*(6-3)*(6-0) + (1/2)*(8-6)*(6-0) / (8-3)*(8-0)
Please note that the exact values of a, b, c, d, e, f, and the probabilities P(|X-Y|≤1) and P(min(X,Y)≤6) will depend on the values of the given intervals [3,8] and [0,x].