Calculating the expectation of function that depends on original random variable and smallest order stat
0
I'm calculating the MSE of a particular statistic and I'm ultimately stuck on deriving the following: $$Eleft[sum_{i=1}^n(X_i-X_{n:1})^2right]$$
More specifically, to calculate the expectation $E[g(x_i, x_{n
;1})],$ would I need to use the joint pdf for $X$ and $X_{n:1}$? I don't believe I know how to derive this. Also, I am assuming that the two arguments for the function $g$ are not independent, which may or may not be correct...
probability probability-distributions expected-value
edited Nov 25 at 14:58
gt6989b
33k22452
asked Nov 25 at 14:56
DavidS
337111
|
show 1 more comment
0
I'm calculating the MSE of a particular statistic and I'm ultimately stuck on deriving the following: $$Eleft[sum_{i=1}^n(X_i-X_{n:1})^2right]$$
More specifically, to calculate the expectation $E[g(x_i, x_{n
;1})],$ would I need to use the joint pdf for $X$ and $X_{n:1}$? I don't believe I know how to derive this. Also, I am assuming that the two arguments for the function $g$ are not independent, which may or may not be correct...
probability probability-distributions expected-value
edited Nov 25 at 14:58
gt6989b
33k22452
asked Nov 25 at 14:56
DavidS
337111
Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
1
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10
|
show 1 more comment
0
0
0
I'm calculating the MSE of a particular statistic and I'm ultimately stuck on deriving the following: $$Eleft[sum_{i=1}^n(X_i-X_{n:1})^2right]$$
More specifically, to calculate the expectation $E[g(x_i, x_{n
;1})],$ would I need to use the joint pdf for $X$ and $X_{n:1}$? I don't believe I know how to derive this. Also, I am assuming that the two arguments for the function $g$ are not independent, which may or may not be correct...
probability probability-distributions expected-value
edited Nov 25 at 14:58
gt6989b
33k22452
asked Nov 25 at 14:56
DavidS
337111
I'm calculating the MSE of a particular statistic and I'm ultimately stuck on deriving the following: $$Eleft[sum_{i=1}^n(X_i-X_{n:1})^2right]$$
More specifically, to calculate the expectation $E[g(x_i, x_{n
;1})],$ would I need to use the joint pdf for $X$ and $X_{n:1}$? I don't believe I know how to derive this. Also, I am assuming that the two arguments for the function $g$ are not independent, which may or may not be correct...
probability probability-distributions expected-value
probability probability-distributions expected-value
edited Nov 25 at 14:58
gt6989b
33k22452
asked Nov 25 at 14:56
DavidS
337111
edited Nov 25 at 14:58
gt6989b
33k22452
asked Nov 25 at 14:56
DavidS
337111
edited Nov 25 at 14:58
gt6989b
33k22452
edited Nov 25 at 14:58
gt6989b
33k22452
edited Nov 25 at 14:58
gt6989b
33k22452
33k22452
asked Nov 25 at 14:56
DavidS
337111
asked Nov 25 at 14:56
DavidS
337111
asked Nov 25 at 14:56
DavidS
337111
337111
Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
1
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10
|
show 1 more comment
Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
1
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10
Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
1
1
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10
|
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Note that $$ mathbb{E}left[(X-Y)^2right] = mathbb{E}left[X^2 + Y^2 - 2XYright] = mathbb{E}left[X^2right] + mathbb{E}left[Y^2right] - 2mathbb{E}left[XYright] $$ so $$ begin{split} mathbb{E}left[sum_{k=1}^n left(X_i-X_{(1)}right)^2right] &= sum_{k=1}^n mathbb{E}left[X_i^2+X_{(1)}^2 - 2X_iX_{(1)}right]\ &= sum_{k=1}^n mathbb{E}left[X_i^2right] + n mathbb{E}left[X_{(1)}^2right] - 2 mathbb{E}left[X_{(1)} sum_{k=1}^n X_iright] end{split} $$ I am not sure if that last term is easier term-by-term or as a whole sum...
– gt6989b
Nov 25 at 15:09
Right, thanks. Still, as you mention, I'm not sure how to deal with the last term...
– DavidS
Nov 25 at 15:15
What is the parent distribution?
– StubbornAtom
Nov 25 at 15:47
1
The last term could be expressed $displaystyle 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{i} right] = 2mathbb{E}left[X_{(1)}sum_{k=1}^n X_{(i)} right] = 2sum_{i=1}^n mathbb{E}[X_{(1)}X_{(i)}]$
– BGM
Nov 25 at 16:32
@StubbornAtom By parent distribution, do you mean the distribution of $X$? The pdf is $f(x)=sigma^{-1}e^{-(x-mu)/sigma}$ where $x>mu$.
– DavidS
Nov 25 at 19:10