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
Let (X_1, Y_1),\ldots , (X_ n, Y_ n)\sim (X,Y) be i.i.d. samples with the given distribution, i.e. X \in [0,1] and Y \in \mathbb {R} such that X \sim \mathsf{Unif}(0,1) and the conditional distribution of Y|X is given by \mathsf{Exp}(1/(a+bX)^2).
Is the maximum likelihood estimator of \beta =\begin{pmatrix} a\\ b\end{pmatrix} the least square estimator? Choose the correct answer with the correct reason.
Yes, since the model distribution of Y|X is in the exponential family.
Yes, since the link function is the canonical link
No, since the link function is not the canonical link
No, since the MLE is NOT obtained by minimizing \displaystyle \sum _ i^ n (Y_ i-X_ i\beta )^2 where \beta =\begin{pmatrix} a\\ b\end{pmatrix}.