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}.
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}.
Let's go through each question step-by-step:
1) What is the regression function m(x) of Y onto X?
To find the regression function, we need to determine the expected value of Y given X. This can be found by integrating Y over its conditional distribution:
m(x) = E(Y|X) = ∫y * g(y|X) dy
Since the conditional distribution of Y|X is given by Exp(1/(a+bX)^2), the integral becomes:
m(x) = ∫y * (1/(a+bX)^2) * e^(-y/(a+bX)^2) dy
To solve this integral, we can use integration by parts. The result is:
m(x) = 1/(a+bx)
Therefore, the correct answer is m(x) = 1/(a+bx).
2) Which of the following is an appropriate choice for a link function for this model?
The link function is used to relate the mean of Y to the linear predictor, which in this case is m(x) = 1/(a+bx).
By looking at the possible choices, none of them directly relates to the function m(x) = 1/(a+bx). Therefore, none of the options given are appropriate choices for the link function.
3) Is the maximum likelihood estimator of β = [a, b] the least square estimator?
To determine this, we need to compare the method of obtaining the MLE and the least square estimator.
The MLE is obtained by maximizing the likelihood function, which is the product of the conditional pdfs of Y given X for the observed data points. In this case, it would involve maximizing the likelihood function with respect to the parameters a and b.
The least square estimator, on the other hand, is obtained by minimizing the sum of squared errors between the observed values of Y and the predicted values based on the linear regression model.
Since the MLE is not obtained by minimizing the sum of squared errors, the correct answer is No, the MLE of β is not the least square estimator.