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.
Which of the following is an appropriate choice for a link function for this model?
\displaystyle h(\mu )=\frac{1}{\mu }
\displaystyle h(\mu )=-\frac{1}{\mu }
\displaystyle h(\mu )=\mu ^2
\displaystyle h(\mu )=\sqrt{\mu }
\displaystyle h(\mu )=\frac{1}{\sqrt{\mu }}
none of the above
The regression function m(x) of Y onto X is given by m(x) = E(Y|X=x).
To find m(x), we need to find the expected value of Y given X=x using the conditional distribution.
E(Y|X=x) = ∫ y * g(y|X=x) dy
Since the conditional pdf of Y given X is given by g(y|X), we can substitute it into the expected value formula:
E(Y|X=x) = ∫ y * \frac{1}{(a+bx)^2} e^{-y/(a+bx)^2} dy
To evaluate this integral, we use the fact that the exponential distribution with parameter λ has expected value 1/λ.
Therefore, we can rewrite the integral as:
E(Y|X=x) = ∫ y * \frac{1}{(a+bx)^2} e^{-y/(a+bx)^2} dy
= (a+bx)^2 * ∫ \frac{y}{(a+bx)^2} * \frac{1}{(a+bx)^2} e^{-y/(a+bx)^2} dy
= (a+bx)^2 * E(Y|X=x)
From the above equation, we can see that the regression function m(x) = (a+bx)^2.
Therefore, the correct answer is m(x) = (a+bx)^2.
None of the above options is the correct regression function.
As for the appropriate choice for a link function, the link function is typically used in generalized linear models to relate the expected value of the response variable to the linear predictor.
In this case, the appropriate choice for a link function is h(μ) = (μ)^2, where μ is the expected value of Y.
Therefore, the correct answer is h(μ) = (μ)^2.
None of the above options is the correct link function.
To find the regression function m(x) of Y onto X, we need to find the conditional expectation of Y given X.
The conditional expectation of Y given X, denoted as E(Y|X=x), can be found by integrating Y times the conditional pdf g(y|X=x) over the range of possible values for Y, which is positive values.
Therefore,
E(Y|X=x) = ∫[0,∞] y g(y|X=x) dy
Let's calculate this integral:
E(Y|X=x) = ∫[0,∞] y * (1/(a+bx)^2) * e^(-y/(a+bx)^2) dy
To simplify the integral, let's substitute u = y/(a+bx)^2, which implies du = (1/(a+bx)^2) dy.
Also, when y = 0, we have u = 0, and when y = ∞, we have u = ∞.
The integral becomes:
E(Y|X=x) = ∫[0,∞] (a+bx)^2 * u * e^(-u) * du
Now, we can solve this integral:
E(Y|X=x) = (a+bx)^2 * ∫[0,∞] u * e^(-u) * du
This integral is equivalent to the expected value of an exponential distribution with parameter 1. The expected value of an exponential distribution with parameter λ is 1/λ.
So, E(Y|X=x) = (a+bx)^2 * 1/1 = (a+bx)^2 = m(x)
Therefore, the regression function of Y onto X is m(x) = (a+bx)^2.
As for the appropriate choice of the link function for this model, the link function h(μ) can be chosen such that h(μ) = g^(-1)(μ), where g(μ) is the conditional expectation function.
In this case, the conditional expectation function is m(x) = (a+bx)^2.
Therefore, the appropriate choice for the link function is h(μ) = μ^2.
So, the correct option is h(μ) = μ^2.