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).
To find the PDF of Y, we need to integrate the conditional PDF fY|X(y|x) over all possible values of X.
For 0 ≤ y ≤ x ≤ 8, the conditional PDF is given as fY|X(y|x) = 1/x.
To find the PDF of Y, we integrate fY|X(y|x) over the range of X from y to 8:
fY(y) = ∫[y,8] fY|X(y|x) dx
= ∫[y,8] 1/x dx
= ln(8/y)
Here, a = 1, b = 8/y, and c = 8. So, the PDF of Y is:
fY(y) = ⎧⎩⎨ln(8/y) y ∈ [d, e]
ln(c/y) y ∈ [e, f]
0 otherwise,
where d, e, and f are to be determined.
Next, we can calculate the probabilities:
1. P(|X - Y| ≤ 1):
This is equivalent to P(Y - 1 ≤ X ≤ Y + 1).
For a given Y, X can range from Y - 1 to Y + 1.
P(|X - Y| ≤ 1) = ∫[Y-1, Y+1] fX(x) dx
Since X is uniformly distributed on [3,8], its PDF is constant over that interval:
fX(x) = 1 / (8 - 3) = 1/5 for 3 ≤ x ≤ 8.
Therefore, P(|X - Y| ≤ 1) = ∫[Y-1, Y+1] (1/5) dx
= (1/5) * [(Y+1) - (Y-1)]
= 2/5
2. P(min(X, Y) ≤ 6):
This is equivalent to P(X ≤ 6 ∪ Y ≤ 6).
P(X ≤ 6) = ∫[3, 6] fX(x) dx
= ∫[3, 6] (1/5) dx
= (1/5) * [(6) - (3)]
= 3/5
P(Y ≤ 6) = 1 - P(Y > 6)
= 1 - P(X > 6, Y > 6)
= 1 - ∫[6, 8] ∫[6, x] fY|X(y|x) dy dx
= 1 - ∫[6, 8] ∫[6, x] (1/x) dy dx
= 1 - ∫[6, 8] [(1/x) * (x - 6)] dx
= 1 - ∫[6, 8] (1 - 6/x) dx
= 1 - [x - 6ln(x)]|[6, 8]
= 1 - [(8 - 6ln(8)) - (6 - 6ln(6))]
= 1 - (8 - 6ln(8) - 6 + 6ln(6))
= 1 - 2 + 6ln(6) - 6ln(8)
= -1 + 6ln(6/8)
= 6ln(3/4)
Therefore, P(min(X, Y) ≤ 6) = P(X ≤ 6 ∪ Y ≤ 6) = P(X ≤ 6) + P(Y ≤ 6) - P(X ≤ 6 ∩ Y ≤ 6)
= 3/5 + 6ln(3/4) - P(X ≤ 6, Y ≤ 6)
P(X ≤ 6, Y ≤ 6) = ∫[3, 6] ∫[0, x] fY|X(y|x) dy dx
= ∫[3, 6] ∫[0, x] (1/x) dy dx
= ∫[3, 6] (1/x) * [x - 0] dx
= ∫[3, 6] 1 dx
= 6 - 3
= 3
P(min(X, Y) ≤ 6) = 3/5 + 6ln(3/4) - 3
= 3/5 + 6ln(3/4) - 3
= 6ln(3/4) - 12/5
To find the PDF fY(y) of Y, we need to consider the range of values y can take. From the conditional PDF fY|X(y|x) given, we have that y is uniformly distributed on the interval [0, x]. Since X is uniformly distributed on the interval [3, 8], Y can take values in the range [0, 3] if X = 3, and [0, 8] if X = 8.
We can break down the PDF fY(y) of Y into two parts, depending on the possible values of y. Let's calculate each part separately:
1. For y ∈ [0, 3]:
P(Y ≤ y) = P(Y ≤ y | X = x) * P(X = x)
= ∫[0,y] 1/x dx * P(X = x) (using a conditional probability formula)
Since X is uniformly distributed on [3, 8], the probability P(X = x) is equal to 1/(8-3) = 1/5. Using the given recall formula, the integral ∫[0,y] 1/x dx is equal to ln(y/3). Therefore, we have:
P(Y ≤ y) = ln(y/3) * (1/5)
Differentiating P(Y ≤ y) with respect to y will give us the PDF fY(y) for y ∈ [0, 3]. So, we have:
fY(y) = d/dy [ln(y/3) * (1/5)]
= (1/5) * (1/(y/3))
= 3/(5y), for y ∈ [0, 3]
2. For y ∈ (3, 8]:
P(Y ≤ y) = P(Y ≤ y | X = x) * P(X = x)
= ∫[0,y] 1/x dx * P(X = x) (using a conditional probability formula)
Again, using the recall formula, the integral ∫[0,y] 1/x dx is equal to ln(y/3). The probability P(X = x) is equal to 1/(8-3) = 1/5 for x ∈ [3, 8]. So, we have:
P(Y ≤ y) = ln(y/3) * (1/5)
Differentiating P(Y ≤ y) with respect to y will give us the PDF fY(y) for y ∈ (3, 8]. Therefore, we have:
fY(y) = d/dy [ln(y/3) * (1/5)]
= (1/5) * (1/(y/3))
= 3/(5y), for y ∈ (3, 8]
Putting both parts together, we have:
fY(y) = ⎧⎩⎨3/(5y) for y ∈ [0, 3]
3/(5y) for y ∈ (3, 8]
0 otherwise
To calculate P(|X-Y| ≤ 1), we need to find the joint distribution of X and Y. Since both X and Y are continuous random variables, we can use the double integral to calculate the probability:
P(|X-Y| ≤ 1) = ∫∫[|x-y| ≤ 1] fX,Y(x, y) dx dy
To calculate the integral, we need the joint PDF fX,Y(x, y). However, we only have the conditional PDF fY|X(y|x).
To calculate P(min(X,Y) ≤ 6), we can use a similar approach by finding the joint distribution of X and Y:
P(min(X,Y) ≤ 6) = ∫∫[min(x,y) ≤ 6] fX,Y(x, y) dx dy
Again, we need the joint PDF fX,Y(x, y) to calculate this integral.
Unfortunately, based on the given information, we do not have enough to determine the joint PDF of X and Y or calculate these probabilities precisely.
To find the PDF of Y, we need to consider the range of possible values for Y and determine the corresponding probability density function.
Given that X is uniformly distributed on the interval [3, 8], the probability density function of X is:
fX(x) = 1/(8-3) = 1/5, for 3 ≤ x ≤ 8
We will consider the cases where Y falls within different intervals with respect to X.
Case 1: 0 ≤ y ≤ 3
In this case, the probability that Y takes a value between 0 and 3 is 1/5, regardless of the value of X. Therefore, fY(y) = 1/5 for 0 ≤ y ≤ 3.
Case 2: 3 < y ≤ 8
In this case, the probability that Y takes a value between 3 and 8 is given by the conditional PDF fY|X(y|x) = 1/x. Since Y is uniformly distributed on the interval [0, X] given X = x, the cumulative distribution function (CDF) of Y is given by integrating the conditional PDF over the range [0, y] with respect to X, and then taking the derivative with respect to y.
F(Y ≤ y) = ∫[0,y] fY|X(t|x) dt
= ∫[0,y] 1/x dt
= ln(x)|[0,y]
= ln(y)
Taking the derivative, we obtain:
fY(y) = d/dy (ln(y))
= 1/y
Therefore, the probability density function of Y is fY(y) = 1/y for 3 < y ≤ 8.
Combining the two cases, we can write the PDF of Y as:
fY(y) = ⎧⎩⎨1/5 0 ≤ y ≤ 3,
⎧⎩⎨1/y 3 < y ≤ 8,
0 otherwise.
To calculate P(|X - Y| ≤ 1), we need to consider the joint distribution of X and Y. The probability can be obtained by integrating the joint PDF over the region defined by |x - y| ≤ 1.
P(|X - Y| ≤ 1) = ∫∫ |x - y| ≤ 1 fX(x)fY(y) dx dy
Integrating over the region defined by |x - y| ≤ 1, we split the integral into two parts: x < y + 1 and y - 1 < x.
P(|X - Y| ≤ 1) = ∫∫ x < y + 1 fX(x)fY(y) dx dy + ∫∫ y - 1 < x fX(x)fY(y) dx dy
Within the first integral, the limits of integration for x are 3 to y + 1, and for y they are 0 to 8.
For the second integral, the limits of integration for x are y - 1 to 8, and for y they are 0 to 8.
P(|X - Y| ≤ 1) = ∫[0,8] ∫[3,y+1] fX(x)fY(y) dx dy + ∫[0,8] ∫[y-1,8] fX(x)fY(y) dx dy
Plugging in the expressions for fX(x) and fY(y) from before, we get:
P(|X - Y| ≤ 1) = ∫[0,8] ∫[3,y+1] (1/5)(1/y) dx dy + ∫[0,8] ∫[y-1,8] (1/5)(1/y) dx dy
Evaluating the integrals, we find:
P(|X - Y| ≤ 1) = ln(9/8) + ln(8/7) + ln(7/6) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1)
To calculate P(min(X,Y) ≤ 6), we need to consider the joint distribution of X and Y again. The probability can be obtained by integrating the joint PDF over the region defined by min(x, y) ≤ 6.
P(min(X, Y) ≤ 6) = ∫∫ min(x, y) ≤ 6 fX(x)fY(y) dx dy
Integrating over the region defined by min(x, y) ≤ 6, we split the integral into two parts: x ≤ 6 and y ≤ 6. This gives:
P(min(X, Y) ≤ 6) = ∫[0,6] ∫[0,y] fX(x)fY(y) dx dy + ∫[6,8] ∫[0,6] fX(x)fY(y) dx dy
Using the expressions for fX(x) and fY(y) from before, we have:
P(min(X, Y) ≤ 6) = ∫[0,6] ∫[0,y] (1/5)(1/y) dx dy + ∫[6,8] ∫[0,6] (1/5)(1/y) dx dy
Evaluating the integrals, we find:
P(min(X, Y) ≤ 6) = ln(7/2) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1)
Therefore, P(|X - Y| ≤ 1) = ln(9/8) + ln(8/7) + ln(7/6) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1) and P(min(X, Y) ≤ 6) = ln(7/2) + ln(6/5) + ln(5/4) + ln(4/3) + ln(3/2) + ln(2/1).