let X \in [0,1] and Y \in \mathbb {R} be two random variables such that X \sim \mathsf{Unif}(0,1) and the conditional distribution of Y|X is given by \mathsf{Exp}(1/(a+bX)^2).
In other words, the conditional pdf of Y given X is
g(y|X)=\left\{ \begin{array}{ll} \frac{1}{(a+bX)^2} e^{-y/(a+bX)^2} & \text {if } y>0\\ 0 & \text {otherwise} \end{array} \right.
(Or in bigger fonts: g(y|X)=\, \frac{1}{(a+bX)^2} \exp \left(-\frac{y}{(a+bX)^2}\right) if y>0 and 0 otherwise.)
For the model above, what is the regression function m(x) of Y onto X?
m(x)= 1/(a+bx)^2
m(x)=(a+bx)^2
m(x)= 1/(a+bx)
m(x)= a+bx
none of the above.
The regression function of Y onto X is denoted as m(x) and it represents the expected value of Y given X=x.
To find the regression function, we need to calculate E(Y|X=x). For a continuous variable, this can be done by integrating the conditional pdf over the range of Y.
In this case, the conditional pdf of Y given X is given by:
g(y|X) = (1/(a+bX)^2) * e^(-y/(a+bX)^2), if y > 0
To find E(Y|X=x), we integrate y * g(y|X) over the range of y, which is from 0 to infinity:
E(Y|X=x) = Integral [y * (1/(a+bx)^2) * e^(-y/(a+bx)^2)] dy (from 0 to infinity)
Performing the integration, we find:
E(Y|X=x) = 1/(a+bx)
Therefore, the regression function m(x) of Y onto X is m(x) = 1/(a+bx).
The correct answer is:
m(x) = 1/(a+bx)
The regression function m(x) of Y onto X is defined as the expected value of Y given X. In other words, it is the average value of Y for a given value of X.
To find the regression function m(x) in this case, we need to calculate E(Y|X=x), the expected value of Y given X=x.
Using the conditional distribution of Y|X, we can calculate the expected value as follows:
E(Y|X=x) = ∫[0,∞] y * g(y|X=x) dy
Let's compute this integral step-by-step.
∫[0,∞] y * g(y|X=x) dy
= ∫[0,∞] y * (1/(a+bx)^2) * exp(-y/(a+bx)^2) dy
To make this integral easier to work with, let's make a substitution u = y/(a+bx)^2.
Since du = (1/(a+bx)^2) dy, we can rewrite the integral as:
∫[0,∞] (a+bx)^2 * u * exp(-u) du
Now, expanding (a+bx)^2, we have:
∫[0,∞] (a^2 + 2abx + b^2x^2) * u * exp(-u) du
To evaluate this integral, we can split it into three separate integrals:
∫[0,∞] a^2 * u * exp(-u) du
+ ∫[0,∞] 2abx * u * exp(-u) du
+ ∫[0,∞] b^2x^2 * u * exp(-u) du
The first integral can be evaluated as a standard integral:
∫[0,∞] a^2 * u * exp(-u) du = a^2 * Γ(2)
where Γ(2) is the gamma function evaluated at 2.
The second and third integrals involve a combination of exponential functions and polynomial terms. These integrals are more complicated and may not have closed-form solutions.
Therefore, in general, the regression function m(x) cannot be simplified to a simple expression. The correct answer is "none of the above."