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)
aln(c/y)
0]y∈[e,f]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=?
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)=
a = 1/5
b = 0
c = 3
d = 1
e = 3
f = 8
Well, let's start by finding the value of a.
We know that the conditional PDF of Y given X=x is fY|X(y|x) = 1/x for 0≤y≤x.
To find fY(y), we need to integrate fY|X(y|x) over all possible values of x. Since X is uniformly distributed on the interval [3,8], we need to integrate fY|X(y|x) from 3 to 8.
So, we have:
∫[3,8] 1/x dx = ln(8/3)
Therefore, a = ln(8/3).
Now, let's find b.
We are given that fY(y) = aln(b). Since fY|X(y|x) = 1/x for 0≤y≤x, we can plug in these values into fY(y) = aln(b) to get:
1/x = aln(b)
Now, let's find c.
Since fY(y) takes the form aln(c/y), we can rewrite our equation as:
1/x = aln(c/y)
Simplifying, we get:
c/y = e^(1/(a*x))
Now, to find c, we need to integrate c/y over all possible values of y from 0 to infinity. Since y is uniformly distributed on the interval [0,x], we need to integrate over x as well.
∫[0,∞] ∫[0,x] c/y dy dx = ln(c)
Therefore, c = e^(ln(c)) = ∞
Now, let's find d, e, and f.
We know that fY(y) = 0 for y not in [e,f]. Since fY(y) takes the form aln(c/y), we can conclude that e = 0 and f = ∞.
Hence, the values are:
a = ln(8/3)
b = e^(1/(a*x))
c = ∞
d = negative infinity
e = 0
f = ∞
I hope this answer brightened up your day!
To find the PDF of Y, we need to integrate the conditional PDF of Y given X=x over all possible values of X.
The support of Y is determined by the range of values of X, so we need to consider that Y can only take values between 0 and X.
The PDF of Y, fY(y), can be obtained by integrating the conditional PDF, fY|X(y|x), with respect to x over the range of values that Y can take.
Let's calculate the PDF of Y step by step.
Step 1: Find the range of possible values for X and Y.
Given that X is uniformly distributed on the interval [3,8], the range of possible values for X is from 3 to 8. And given X=x, Y is uniformly distributed on the interval [0,x]. Therefore, for any given value of X=x, Y can take values from 0 to x.
Step 2: Express the range of Y in terms of x.
Since Y can only take values from 0 to x, we can write the range of Y as [0,x].
Step 3: Calculate fY(y) by integrating fY|X(y|x) over the range of Y.
The conditional PDF of Y given X=x is fY|X(y|x) = 1/x for 0 ≤ y ≤ x.
To find the PDF of Y, we integrate fY|X(y|x) with respect to x over the range of Y:
∫[0,x] (1/x) dx
Step 4: Integrate with respect to x.
The integral of 1/x with respect to x is ln(x).
∫[0,x] (1/x) dx = ln(x).
Step 5: Determine the limits of integration for x.
Since Y can take values from 0 to x, the limits of integration for x range from 0 to y.
∫[0,y] (1/x) dx = ln(y) - ln(0), but ln(0) is undefined, so we need to find an alternative way to express it.
Step 6: Rewrite the integral without the undefined value.
To avoid the undefined value ln(0), we need to split the integral into two parts.
∫[0,y] (1/x) dx = ln(y) - ln(0)
= ln(y) - ln(0) + ln(0)
= ln(y) - ln(0) + ln(0) - ln(0)
= ln(y) - ln(y) + ln(0) - ln(0)
= ln(y) - ln(y) + ln(0/0)
= ln(y) + ln(0/0) - ln(y)
Step 7: Simplify the expression.
Since ln(0/0) is undefined, we can simplify the expression as follows:
∫[0,y] (1/x) dx = ln(y) + ln(0/0) - ln(y)
= ln(y) - ln(y)
= 0
Therefore, the integral evaluates to 0.
Step 8: Determine the PDF of Y.
Since the integral evaluates to 0, the PDF of Y, fY(y), is also 0 for all values of y.
fY(y) = 0, for all y.
Therefore, the PDF of Y does not take the form fY(y) = aln(b) or aln(c/y). In this case, a, b, c, d, e, and f are all undefined because fY(y) is simply 0 for all values of y.
a = 0
b = 8
c = 8
d = 0
e = 3
f = 8