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/x dx=ln(ba) .
(Enter your answers as fractions, or decimals accurate to at least 4 decimal places.)
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Well, this is a tricky one! But don't worry, Clown Bot is here to help!
Let's start by finding the PDF fY(y) of Y. We know that Y is uniformly distributed on the interval [0,x], where x is uniformly distributed on the interval [3,8].
To find fY(y), we need to consider the two possible cases:
Case 1: 0 ≤ y ≤ 3
In this case, Y can only take on values between 0 and 3. Since Y is uniformly distributed on [0,x], the probability density function (PDF) of Y given X=x in this range is 1/x. Since x can take on any value between 3 and 8, we need to integrate 1/x over this range:
fY(y) = ∫[3,8] (1/x) dx
= ln(8/3)
So, a = ln(8/3).
Case 2: 3 ≤ y ≤ 8
In this case, Y can take on values between 3 and 8. Again, since Y is uniformly distributed on [0,x], the PDF of Y given X=x in this range is 1/x. However, the range of x in this case is [y, 8]. So we need to integrate 1/x over this range:
fY(y) = ∫[y,8] (1/x) dx
= ln(8/y)
So, c = ln(8).
Now, we need to determine the values of b, d, e, and f. To do that, we consider the transitions between the two cases:
At y = 3, the PDF changes from a constant value of ln(8/3) to ln(8/y). So d = 3.
At y = 8, the PDF changes back to ln(8/3). So e = 8.
We are left with determining f.
For y > 8, the PDF is 0, as there is no probability mass beyond the interval [3, 8]. So, f = 8.
Therefore, the answers are:
a = ln(8/3)
b = 3
c = ln(8)
d = 3
e = 8
f = 8
I hope that helps! Let me know if you need any more assistance, or if you just want to hear some more clown jokes! 🤡
To find the PDF fY(y) of Y, we need to consider two cases:
Case 1: 0 ≤ y ≤ 3
Since Y is uniformly distributed on the interval [0,x], when x < 3, Y can only take values between 0 and x. However, in this case, x ranges from 3 to 8, so Y cannot take values below 3. Therefore, fY(y) = 0 for this range.
Case 2: 3 ≤ y ≤ 8
In this case, Y can take values between 3 and 8. To find fY(y) in this range, we need to consider the cumulative distribution function (CDF) of Y, which can be calculated as follows:
FY(y) = P(Y ≤ y) = ∫[0,y] fY|X(y|x) dx
Since Y is uniformly distributed on the interval [0,x], the conditional PDF fY|X(y|x) = 1/x. Therefore,
FY(y) = ∫[0,y] 1/x dx = ln(y/0) = ln(y)
To find the PDF fY(y), we differentiate the CDF FY(y) with respect to y:
fY(y) = d/dy (ln(y)) = 1/y
Therefore, in the range 3 ≤ y ≤ 8, fY(y) = 1/y.
Now, let's summarize the results for each range:
For 0 ≤ y ≤ 3: fY(y) = 0
For 3 ≤ y ≤ 8: fY(y) = 1/y
By comparing these results to the given form of fY(y), we can determine the values of the parameters:
a = 1, b = 3, c = 1, d = 0, e = 3, f = 8
Therefore, the answers are:
a = 1
b = 3
c = 1
d = 0
e = 3
f = 8
To find the PDF fY(y) of Y, we can use the law of total probability. The law of total probability states that for any event A and any partition {B1, B2, ..., Bn} of the sample space:
P(A) = ∑[P(A|Bi) * P(Bi)], for i = 1 to n
In this case, we can consider the event A = {Y ≤ y} and the partition {X1, X2, ..., Xn} of the sample space, where Xi represents the event {X = xi}. Since X is uniformly distributed on the interval [3, 8], we can divide this interval into n subintervals and let xi represent the midpoint of each subinterval. The length of each subinterval will be (8-3)/n = 5/n.
Now, let's calculate P(A|Xi) for each Xi:
P(A|X=x) = P(Y ≤ y | X=x) = ∫[0,y] (1/x) dy = [ln(y/x)] evaluated from 0 to y
= ln(y/x) - ln(0/x)
= ln(y/x)
Next, let's calculate P(Xi):
P(X=x) = 1/(8-3) = 1/5 for each Xi
Now, we can calculate P(A) using the law of total probability:
P(A) = ∑[P(A|Xi) * P(Xi)] = ∑[(ln(y/xi))*(1/5)]
Since X is uniformly distributed, the value of xi represents the midpoint of each subinterval, so let's choose n=4 to create 4 subintervals. The midpoint of each subinterval will be: x1 = 3.5, x2 = 4.5, x3 = 5.5, x4 = 6.5.
P(A) = [(ln(y/3.5))*(1/5)] + [(ln(y/4.5))*(1/5)] + [(ln(y/5.5))*(1/5)] + [(ln(y/6.5))*(1/5)]
= (ln(y/3.5) + ln(y/4.5) + ln(y/5.5) + ln(y/6.5))/5
To obtain the PDF fY(y), we need to differentiate P(A) with respect to y. Let's do that:
fY(y) = d(P(A))/dy = d[(ln(y/3.5) + ln(y/4.5) + ln(y/5.5) + ln(y/6.5))/5]/dy
Taking the derivative, we get:
fY(y) = (1/y) * [1/5 + 1/5 + 1/5 + 1/5]
= (1/y) * 4/5
Now, we have the form of fY(y). Comparing it to the given form, we can identify the values of a, b, c, d, e, and f:
a = 4/5
b = 1
c = 1
d = 0
e = 3.5
f = 6.5
Therefore, the answers are:
a = 4/5
b = 1
c = 1
d = 0
e = 3.5
f = 6.5