The random variable X has a PDF of the form
fx(x)=1/x^2 for x>1, 0 otherwise
let Y = X^2 for y>1 the PDF of Y it takes the form fy(y)= a/y^b. Find the values of a and b
a=1/2
b=3/2
Why did the variable go to therapy?
Because it had an identity crisis!
To find the PDF of Y, we need to use the transformation method.
Let's start by finding the cumulative distribution function (CDF) of Y, denoted by FY(y). The CDF is defined as the probability that Y is less than or equal to y.
FY(y) = P(Y ≤ y)
Since Y = X^2, we can rewrite the equation as:
FY(y) = P(X^2 ≤ y)
Now, let's determine the values of X that satisfy this inequality.
X^2 ≤ y
Taking the square root of both sides (note that X is positive):
X ≤ √y
From the PDF of X, we know that P(X > x) = 0 for x ≤ 1. Therefore, we can set the lower limit of integration as 1:
FY(y) = ∫[1,√y] fx(x) dx
Substituting the PDF of X into the integral:
FY(y) = ∫[1,√y] 1/x^2 dx
Using the integral, we can integrate the above expression:
FY(y) = [-1/x] evaluated from 1 to √y
= -(1/√y) + 1
Now, let's find the PDF of Y by differentiating the CDF with respect to y:
fy(y) = d(FY(y))/dy
fy(y) = d/dy (-(1/√y) + 1)
= 1/(2y^(3/2))
Comparing the derived PDF fy(y) with the general form a/y^b, we can see that a = 1 and b = 3/2.
Therefore, the values of a and b are a = 1 and b = 3/2 for the PDF of Y.
To find the values of a and b for the PDF of Y, we need to consider the transformation from X to Y.
Given that Y = X^2, we can determine the relationship between the PDFs of X and Y by finding the derivative of the CDF (cumulative distribution function) of Y with respect to y.
Let's proceed step by step:
1. Determine the CDF of Y:
The CDF of Y, denoted by FY(y), can be calculated by substituting Y = X^2 into the CDF of X, denoted by FX(x), and solving for x in terms of y. Since x > 1, we have:
FY(y) = P(Y ≤ y)
= P(X^2 ≤ y)
= P(X ≤ √y)
= FX(√y)
= ∫[1, √y] fx(x) dx
2. Differentiate FY(y) with respect to y:
To find the PDF of Y, we differentiate the CDF FY(y) with respect to y:
fy(y) = d/dy [FY(y)]
= d/dy [∫[1, √y] fx(x) dx]
Now, we can differentiate the integral with respect to y. Applying the Fundamental Theorem of Calculus, we have:
fy(y) = d/dy [∫[1, √y] fx(x) dx]
= fx(√y) * d(√y)/dy
3. Calculate d(√y)/dy:
To find d(√y)/dy, we apply the chain rule of differentiation:
d(√y)/dy = (1/2) * (1/√y)
= 1/(2√y)
4. Substitute fx(√y) and d(√y)/dy into fy(y):
Substituting fx(√y) = 1/(√y)^2 = 1/y and d(√y)/dy = 1/(2√y) into fy(y), we get:
fy(y) = fx(√y) * d(√y)/dy
= (1/y) * (1/(2√y))
= 1/(2y√y)
Comparing this form to the given form fy(y) = a/y^b, we have:
a = 1/2
b = 3
Therefore, the values of a and b for the PDF of Y are a = 1/2 and b = 3.