Question
When computing the least-squares estimator, we are computing some \hat{{\boldsymbol \beta }} which minimizes the error
\min _{{\boldsymbol \beta }\in \mathbb {R}^ p} \| \mathbf Y- \mathbb {X}{\boldsymbol \beta }\| _2^2
where \| v\| _2 is the Euclidean norm.
Let n be the number of samples and let each \mathbf X_ i be p-dimensional. (For example, n might be the number of patients, and p-1 is the number of covariates that we are trying to study - e.g. height, weight, age and blood pressure as in the previous problem.)
Recall that by employing the same technique of computing the gradient (with respect to the components of {\boldsymbol \beta }) and setting it equal to zero, we can show that \hat{{\boldsymbol \beta }} must satisfy the score equation
\mathbb {X}^ T \mathbb {X} \hat{{\boldsymbol \beta }} = \mathbb {X}^ T \mathbf Y.
We would like to isolate \hat{{\boldsymbol \beta }} by multiplying by (\mathbb {X}^ T \mathbb {X})^{-1} from the left. Which of the following conditions, each on its own, guarantees that \mathbb {X}^ T \mathbb {X} is invertible? Choose all that apply.
Hint: Use the fact from linear algebra that \mathrm{rank}(A^ T A) = \mathrm{rank}(A). What are the dimensions of \mathbb {X}^ T \mathbb {X}?
There are at least as many observations as covariates (i.e. n \geq p).
There are at least as many covariates as observations (i.e. n \leq p).
\mathbb {X} has rank n, where n is the number of samples.
\mathbb {X} has rank p, where p is the number of covariates.
There are at least p distinct samples amongst the X's, so that \mathbb {X} has at least p distinct rows.
There are at least p distinct values amongst the Y.
\min _{{\boldsymbol \beta }\in \mathbb {R}^ p} \| \mathbf Y- \mathbb {X}{\boldsymbol \beta }\| _2^2
where \| v\| _2 is the Euclidean norm.
Let n be the number of samples and let each \mathbf X_ i be p-dimensional. (For example, n might be the number of patients, and p-1 is the number of covariates that we are trying to study - e.g. height, weight, age and blood pressure as in the previous problem.)
Recall that by employing the same technique of computing the gradient (with respect to the components of {\boldsymbol \beta }) and setting it equal to zero, we can show that \hat{{\boldsymbol \beta }} must satisfy the score equation
\mathbb {X}^ T \mathbb {X} \hat{{\boldsymbol \beta }} = \mathbb {X}^ T \mathbf Y.
We would like to isolate \hat{{\boldsymbol \beta }} by multiplying by (\mathbb {X}^ T \mathbb {X})^{-1} from the left. Which of the following conditions, each on its own, guarantees that \mathbb {X}^ T \mathbb {X} is invertible? Choose all that apply.
Hint: Use the fact from linear algebra that \mathrm{rank}(A^ T A) = \mathrm{rank}(A). What are the dimensions of \mathbb {X}^ T \mathbb {X}?
There are at least as many observations as covariates (i.e. n \geq p).
There are at least as many covariates as observations (i.e. n \leq p).
\mathbb {X} has rank n, where n is the number of samples.
\mathbb {X} has rank p, where p is the number of covariates.
There are at least p distinct samples amongst the X's, so that \mathbb {X} has at least p distinct rows.
There are at least p distinct values amongst the Y.
Answers
There are no human answers yet.
Answered by
GPT 3.5
The conditions that guarantee that \mathbb {X}^ T \mathbb {X} is invertible are:
1) There are at least as many observations as covariates (i.e. n \geq p).
2) \mathbb {X} has rank p, where p is the number of covariates.
These conditions ensure that the matrix \mathbb {X}^ T \mathbb {X} has full rank and is therefore invertible.
1) There are at least as many observations as covariates (i.e. n \geq p).
2) \mathbb {X} has rank p, where p is the number of covariates.
These conditions ensure that the matrix \mathbb {X}^ T \mathbb {X} has full rank and is therefore invertible.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.