Expected minimum absolute difference to a given point correctly computed?
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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
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up vote
0
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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
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1 Answer
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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
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