Model Selection (k-piece-constant function)
$begingroup$
A $k$-piece-constant function is define by $k-1$ thresholds $-100<t_1<t_2<......<t_{k-1}<100$ and $k$ values as $a_1,a_2,......,a_k$
The function is defined as follows-
If $x<t_1$ then $f(x)=a_1$
If $t_1<x<t_2$ then $f(x)=a_2$
If $t_2<x<t_3$ then $f(x)=a_3$
.
.
.
If $t_{i-1}<x<t_i$ then $f(x)=a_i$
.
.
.
If $t_{k-1}<x<t_k$ then $f(x)=a_k$
Let $f$ be a -piece-constant function. Suppose you are given $n$ data points $((x_1,y_1),(x_2,y_2),.......,(x_n,y_n))$ each of which is generated in the following way:
1. first, $x$ is drawn according to the uniform distribution over the range $[-100,100].$
2. second $y$ is chosen to be $f(x)+omega$ where $omega$ is drawn according to the normal distribution $N(mu,sigma^2)$
You partition the data into a training set and a test set of equal sizes. For each $j=1,2,...$ you find the $j$ -piece-constant function $g_j$ that minimizes the root-mean-square-error on the training set. Denote by $train(j)$ the RMSE on the training set and by $test(j)$ the RMSE on the test set.
Which of the following statements is correct?
$train(j)$ is a monotonically non-increasing function
$test(j)$ is a monotonically non-increasing function
$test(j)$ has a minimum close to $j=k$
$train(j)$ has a minimum close to $j=k$
if $j>n/2$ $train(j)=0$
I have absolutely no clue of the question. I somewhat understood the k-piece-function (which is perhaps new to me) but still unable to crack the question.
P.S. I know the idea behind training and testing sets
statistics regression
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
$endgroup$
add a comment |
$begingroup$
A $k$-piece-constant function is define by $k-1$ thresholds $-100<t_1<t_2<......<t_{k-1}<100$ and $k$ values as $a_1,a_2,......,a_k$
The function is defined as follows-
If $x<t_1$ then $f(x)=a_1$
If $t_1<x<t_2$ then $f(x)=a_2$
If $t_2<x<t_3$ then $f(x)=a_3$
.
.
.
If $t_{i-1}<x<t_i$ then $f(x)=a_i$
.
.
.
If $t_{k-1}<x<t_k$ then $f(x)=a_k$
Let $f$ be a -piece-constant function. Suppose you are given $n$ data points $((x_1,y_1),(x_2,y_2),.......,(x_n,y_n))$ each of which is generated in the following way:
1. first, $x$ is drawn according to the uniform distribution over the range $[-100,100].$
2. second $y$ is chosen to be $f(x)+omega$ where $omega$ is drawn according to the normal distribution $N(mu,sigma^2)$
You partition the data into a training set and a test set of equal sizes. For each $j=1,2,...$ you find the $j$ -piece-constant function $g_j$ that minimizes the root-mean-square-error on the training set. Denote by $train(j)$ the RMSE on the training set and by $test(j)$ the RMSE on the test set.
Which of the following statements is correct?
$train(j)$ is a monotonically non-increasing function
$test(j)$ is a monotonically non-increasing function
$test(j)$ has a minimum close to $j=k$
$train(j)$ has a minimum close to $j=k$
if $j>n/2$ $train(j)=0$
I have absolutely no clue of the question. I somewhat understood the k-piece-function (which is perhaps new to me) but still unable to crack the question.
P.S. I know the idea behind training and testing sets
statistics regression
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
$endgroup$
add a comment |
$begingroup$
A $k$-piece-constant function is define by $k-1$ thresholds $-100<t_1<t_2<......<t_{k-1}<100$ and $k$ values as $a_1,a_2,......,a_k$
The function is defined as follows-
If $x<t_1$ then $f(x)=a_1$
If $t_1<x<t_2$ then $f(x)=a_2$
If $t_2<x<t_3$ then $f(x)=a_3$
.
.
.
If $t_{i-1}<x<t_i$ then $f(x)=a_i$
.
.
.
If $t_{k-1}<x<t_k$ then $f(x)=a_k$
Let $f$ be a -piece-constant function. Suppose you are given $n$ data points $((x_1,y_1),(x_2,y_2),.......,(x_n,y_n))$ each of which is generated in the following way:
1. first, $x$ is drawn according to the uniform distribution over the range $[-100,100].$
2. second $y$ is chosen to be $f(x)+omega$ where $omega$ is drawn according to the normal distribution $N(mu,sigma^2)$
You partition the data into a training set and a test set of equal sizes. For each $j=1,2,...$ you find the $j$ -piece-constant function $g_j$ that minimizes the root-mean-square-error on the training set. Denote by $train(j)$ the RMSE on the training set and by $test(j)$ the RMSE on the test set.
Which of the following statements is correct?
$train(j)$ is a monotonically non-increasing function
$test(j)$ is a monotonically non-increasing function
$test(j)$ has a minimum close to $j=k$
$train(j)$ has a minimum close to $j=k$
if $j>n/2$ $train(j)=0$
I have absolutely no clue of the question. I somewhat understood the k-piece-function (which is perhaps new to me) but still unable to crack the question.
P.S. I know the idea behind training and testing sets
statistics regression
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
$endgroup$
A $k$-piece-constant function is define by $k-1$ thresholds $-100<t_1<t_2<......<t_{k-1}<100$ and $k$ values as $a_1,a_2,......,a_k$
The function is defined as follows-
If $x<t_1$ then $f(x)=a_1$
If $t_1<x<t_2$ then $f(x)=a_2$
If $t_2<x<t_3$ then $f(x)=a_3$
.
.
.
If $t_{i-1}<x<t_i$ then $f(x)=a_i$
.
.
.
If $t_{k-1}<x<t_k$ then $f(x)=a_k$
Let $f$ be a -piece-constant function. Suppose you are given $n$ data points $((x_1,y_1),(x_2,y_2),.......,(x_n,y_n))$ each of which is generated in the following way:
1. first, $x$ is drawn according to the uniform distribution over the range $[-100,100].$
2. second $y$ is chosen to be $f(x)+omega$ where $omega$ is drawn according to the normal distribution $N(mu,sigma^2)$
You partition the data into a training set and a test set of equal sizes. For each $j=1,2,...$ you find the $j$ -piece-constant function $g_j$ that minimizes the root-mean-square-error on the training set. Denote by $train(j)$ the RMSE on the training set and by $test(j)$ the RMSE on the test set.
Which of the following statements is correct?
$train(j)$ is a monotonically non-increasing function
$test(j)$ is a monotonically non-increasing function
$test(j)$ has a minimum close to $j=k$
$train(j)$ has a minimum close to $j=k$
if $j>n/2$ $train(j)=0$
I have absolutely no clue of the question. I somewhat understood the k-piece-function (which is perhaps new to me) but still unable to crack the question.
P.S. I know the idea behind training and testing sets
statistics regression
statistics regression
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
asked Dec 14 '18 at 12:13
Kriti AroraKriti Arora
396
396
add a comment |
add a comment |
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