Suppose that X is a continuous random variable and that Y=X^4. Then, for y≥0 we have
fY(y)=ay^bfX(−cy^d)+ay^bfX(cy^d),
for some a, b, d, and some c>0. Find a, b, c, and d.
a=1/4
b=-3/4
c=1
d=1/4
Why did the variable go to the psychologist?
Because it had too many "Y"s to deal with! But don't worry, I'm here to help. Let's solve this together, shall we?
First, let's figure out the relationship between Y and X. We know that Y = X^4.
To find the probability density function (PDF) of Y, fY(y), we need to use the transformation technique.
The transformation technique states that if Y = g(X), then the PDF of Y is given by:
fY(y) = fX(g^(-1)(y)) * |(dg^(-1)(y))/(dy)|
In our case, g(X) = X^4. So, we need to find g^(-1)(y) and (dg^(-1)(y))/(dy).
To find g^(-1)(y), we solve for X in terms of Y by taking the fourth root:
g^(-1)(y) = (y)^(1/4)
Next, we need to find (dg^(-1)(y))/(dy):
(dg^(-1)(y))/(dy) = (1/4)*(y)^(-3/4)
Now, we can substitute these results back into the formula for fY(y):
fY(y) = fX((y)^(1/4)) * |(1/4)*(y)^(-3/4)|
So, we have:
fY(y) = a*(y)^(1/4)*b*fX((y)^(1/4))/(4*(y)^3/4)
Now, we need to find the values of a, b, c, and d in the equation provided:
fY(y) = ay^bfX(−cy^d)+ay^bfX(cy^d)
Comparing this equation with what we found, we can see that c = 1/4, d = 3/4, a = b = 1.
Therefore, the values of a, b, c, and d are:
a = 1
b = 1
c = 1/4
d = 3/4
Voila! We got our answer. I hope this explanation wasn't too "random" for you!
To find the values of a, b, c, and d, we need to use the probability density function (pdf) transformation formula for continuous random variables.
In this case, let's start by finding the pdf of Y, denoted as fY(y), using the transformation formula:
fY(y) = |dx/dy| * fX(x),
where dx/dy is the derivative of X with respect to Y.
Given that Y = X^4, we can find the derivative as:
dy/dx = 4x^3.
Rearranging the equation, we get:
dx/dy = 1 / (4x^3).
Now, let's substitute this into the formula for fY(y):
fY(y) = |1 / (4x^3)| * fX(x).
Since y ≥ 0 and Y = X^4, we have x = y^(1/4), where x is the range of X.
Note that Y = 0 when X = 0.
Plugging in the x value, we get:
fY(y) = |1 / (4(y^(1/4))^3)| * fX(y^(1/4))
Simplifying:
fY(y) = 1 / (4y^(3/4)) * fX(y^(1/4)).
Comparing this with the given expression for fY(y):
fY(y) = ay^bfX(-cy^d) + ay^bfX(cy^d),
we can conclude that:
a = 1 / (4y^(3/4)),
b = 1,
c = 1,
and d = 1/4.
Therefore, the values of a, b, c, and d are:
a = 1 / (4y^(3/4)),
b = 1,
c = 1,
d = 1/4.
To find the values of a, b, c, and d, we need to use the transformation technique for probability density functions (PDFs) of random variables.
Given that Y=X^4 and X is a continuous random variable, we can use the transformation technique to find the PDF of Y.
The formula for the PDF of Y, denoted as fY(y), is given by:
fY(y) = |fX(g(y)).(d/dy)(g(y))|
where fX(x) is the PDF of X and g(y) is the inverse function of Y. In this case, Y=X^4, so g(y) would be the fourth root of y.
Now, let's find the values of a, b, c, and d.
Step 1: Calculate g(y) by taking the fourth root of y:
g(y) = y^(1/4)
Step 2: Find the derivative of g(y) with respect to y:
(d/dy)(g(y)) = (1/4)y^(-3/4)
Step 3: Plug in g(y) and (d/dy)(g(y)) into the formula for fY(y):
fY(y) = |fX(y^(1/4)).(1/4)y^(-3/4)|
Given the form of fY(y) in the question:
fY(y) = ay^bfX(−cy^d) + ay^bfX(cy^d)
We can compare the two equations and derive the values of a, b, c, and d.
Comparing the functions, we can see that:
a = |fX(y^(1/4)).(1/4)|
b = 1
c = 1
d = 3/4
Therefore, the values of a, b, c, and d are:
a = |fX(y^(1/4)).(1/4)|
b = 1
c = 1
d = 3/4
Note that the exact values of a would depend on the specific PDF of X, which is not provided in the question. However, a can be determined by calculating fX(y^(1/4)) for a given value of y.