Expected minimum absolute difference to a given point correctly computed?
up vote
0
down vote
favorite
I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:
$$
mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
$$
I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.
Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?
If not, how would I go about this problem?
probability probability-theory probability-distributions
add a comment |
up vote
0
down vote
favorite
I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:
$$
mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
$$
I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.
Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?
If not, how would I go about this problem?
probability probability-theory probability-distributions
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:
$$
mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
$$
I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.
Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?
If not, how would I go about this problem?
probability probability-theory probability-distributions
I am trying to compute the expected absolute difference between $n$ iid uniformly between $[- epsilon, epsilon]$ sampled points to a given point $- epsilon< c < epsilon$:
$$
mathrm{E}(mathrm{min}(|x_1-c|,|x_2-c|,...,|x_n-c|))
$$
I know, that the expected minimum of n iid uniformly between $[0, epsilon]$ sampled points is $frac{epsilon}{n+1}$.
Can I simply put my coordinate systems origin onto c and say I have an expected minimum value of $frac{epsilon - c}{n+1}$ on the right, and $frac{epsilon + c}{n+1}$ on the left, and for the overall distribution I have an weighted average (wheigted with the width, so $epsilon-c$ and $epsilon+c$ respectively) of $frac{1}{2 epsilon} cdotfrac{(epsilon-c)^2 + (epsilon+c)^2}{n+1}$?
If not, how would I go about this problem?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Nov 16 at 0:04
Luca Thiede
1486
1486
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:
$$ begin{align}
S(x)&=Pr{|X_i - c| > x} \
&= Pr{X_i > c + x} + Pr{X_i < c - x} \
&= begin{cases}
1 & text {if} & x leq 0 \
displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
0 & text{if} & x geq epsilon + |c|
end{cases}
end{align}$$
Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
$$ begin{align}
E[min|X_i - c|]
&= int_0^{+infty}S(x)^ndx \
&= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
+ int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
&= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
&= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
&= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
end{align}$$
As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:
$$ begin{align}
S(x)&=Pr{|X_i - c| > x} \
&= Pr{X_i > c + x} + Pr{X_i < c - x} \
&= begin{cases}
1 & text {if} & x leq 0 \
displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
0 & text{if} & x geq epsilon + |c|
end{cases}
end{align}$$
Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
$$ begin{align}
E[min|X_i - c|]
&= int_0^{+infty}S(x)^ndx \
&= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
+ int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
&= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
&= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
&= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
end{align}$$
As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.
add a comment |
up vote
1
down vote
accepted
Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:
$$ begin{align}
S(x)&=Pr{|X_i - c| > x} \
&= Pr{X_i > c + x} + Pr{X_i < c - x} \
&= begin{cases}
1 & text {if} & x leq 0 \
displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
0 & text{if} & x geq epsilon + |c|
end{cases}
end{align}$$
Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
$$ begin{align}
E[min|X_i - c|]
&= int_0^{+infty}S(x)^ndx \
&= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
+ int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
&= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
&= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
&= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
end{align}$$
As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:
$$ begin{align}
S(x)&=Pr{|X_i - c| > x} \
&= Pr{X_i > c + x} + Pr{X_i < c - x} \
&= begin{cases}
1 & text {if} & x leq 0 \
displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
0 & text{if} & x geq epsilon + |c|
end{cases}
end{align}$$
Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
$$ begin{align}
E[min|X_i - c|]
&= int_0^{+infty}S(x)^ndx \
&= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
+ int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
&= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
&= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
&= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
end{align}$$
As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.
Since $min|X_i - c|$ is non-negative and you want to compute its expectation, we can compute its survival function first. Consider the survival function of individual $|X_i - c|$:
$$ begin{align}
S(x)&=Pr{|X_i - c| > x} \
&= Pr{X_i > c + x} + Pr{X_i < c - x} \
&= begin{cases}
1 & text {if} & x leq 0 \
displaystyle 1 - frac {x} {epsilon} & text{if} & 0 < x leq epsilon - |c|\
displaystyle frac {epsilon + |c| - x} {2epsilon} & text {if} & epsilon - |c| < x < epsilon + |c| \
0 & text{if} & x geq epsilon + |c|
end{cases}
end{align}$$
Then the survival function of $min |X_i - c|$ will be $S(x)^n$ and thus
$$ begin{align}
E[min|X_i - c|]
&= int_0^{+infty}S(x)^ndx \
&= int_0^{epsilon - |c|} left(1 - frac {x} {epsilon} right)^n dx
+ int_{epsilon - |c|}^{epsilon + |c|} left(frac {epsilon + |c| - x} {2epsilon} right)^n dx \
&= left. frac {-epsilon} {n+1}left(1 - frac {x} {epsilon} right)^{n+1} right|_0^{epsilon - |c|} + left.frac {-2epsilon} {n+1} left(frac {epsilon + |c| - x} {2epsilon} right)^{n+1} right|_{epsilon - |c|}^{epsilon + |c|} \
&= frac {epsilon} {n+1}left(1 - frac {|c|^{n+1}} {epsilon^{n+1}} + 2frac {|c|^{n+1}} {epsilon^{n+1}}right) \
&= frac {epsilon} {n+1}left(1 + frac {|c|^{n+1}} {epsilon^{n+1}} right)
end{align}$$
As you see, the answer is minimized when $c = 0$ and it reduced back to the previous solution.
answered Nov 16 at 9:22
BGM
3,715148
3,715148
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000520%2fexpected-minimum-absolute-difference-to-a-given-point-correctly-computed%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown