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The Bayes predictor of the square loss is $Bbb E_P[Ymid X=x]$?

0

$begingroup$

Let $(X,Y) in Bbb X times Bbb Y$ be jointly distributed according
to distribution $P$. Let $h: Bbb X rightarrow tilde Bbb Y$,
where $tilde Bbb Y$ is a predicted output. $ $Let $L(h,P) equiv
Bbb E_P[l(Y, h(X))]$ where $l$ is some loss function.

Show that $f = arg min_h L(h,P) = Bbb E_p[Y mid X = x]$ if $l$ is the
square loss function: $l(Y, h(X)) = (y - h(x))^2$

I figured I show this by showing any other $h$ leads to a larger $L(h,P)$ than $Bbb E_P[Ymid X=x]$.

I start with $$Bbb E_P[(y - Bbb E_p[Ymid X=x])^2] le Bbb E_P[(y - h(x))^2]$$

Then expanding we have:

$$Bbb E_P[y^2-2yBbb E_p[Y|X=x] + Bbb E_P[Ymid X=x]^2] le Bbb E_P[y^2 - 2yh(x) + h(x)^2]$$

And simplifying:

$$-2Bbb E_P[y]Bbb E_p[Ymid X=x] + Bbb E_P[Ymid X=x]^2 le -2Bbb E_P[yh(x)] + Bbb E_P[h(x)^2]$$

But from here I'm a little stuck as to how to continue.

Does anyone have any ideas?

probability machine-learning

share|cite|improve this question

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You want to show that the conditional expectation minimises the square loss. You can find discussion of this here: stats.stackexchange.com/questions/71863/….
  $endgroup$
  – Minus One-Twelfth
  Mar 17 at 14:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "I start with" You don't start with your desired conclusion. You start with what you know.
  $endgroup$
  – leonbloy
  Mar 25 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The link given by MinusOne-Twelfth has effectively answered the question in details. Is there anything else you'd like to know?
  $endgroup$
  – Saad
  Mar 26 at 1:58
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $(X,Y) in Bbb X times Bbb Y$ be jointly distributed according
to distribution $P$. Let $h: Bbb X rightarrow tilde Bbb Y$,
where $tilde Bbb Y$ is a predicted output. $ $Let $L(h,P) equiv
Bbb E_P[l(Y, h(X))]$ where $l$ is some loss function.

Show that $f = arg min_h L(h,P) = Bbb E_p[Y mid X = x]$ if $l$ is the
square loss function: $l(Y, h(X)) = (y - h(x))^2$

I figured I show this by showing any other $h$ leads to a larger $L(h,P)$ than $Bbb E_P[Ymid X=x]$.

I start with $$Bbb E_P[(y - Bbb E_p[Ymid X=x])^2] le Bbb E_P[(y - h(x))^2]$$

Then expanding we have:

$$Bbb E_P[y^2-2yBbb E_p[Y|X=x] + Bbb E_P[Ymid X=x]^2] le Bbb E_P[y^2 - 2yh(x) + h(x)^2]$$

And simplifying:

$$-2Bbb E_P[y]Bbb E_p[Ymid X=x] + Bbb E_P[Ymid X=x]^2 le -2Bbb E_P[yh(x)] + Bbb E_P[h(x)^2]$$

But from here I'm a little stuck as to how to continue.

Does anyone have any ideas?

probability machine-learning

share|cite|improve this question

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You want to show that the conditional expectation minimises the square loss. You can find discussion of this here: stats.stackexchange.com/questions/71863/….
  $endgroup$
  – Minus One-Twelfth
  Mar 17 at 14:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "I start with" You don't start with your desired conclusion. You start with what you know.
  $endgroup$
  – leonbloy
  Mar 25 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The link given by MinusOne-Twelfth has effectively answered the question in details. Is there anything else you'd like to know?
  $endgroup$
  – Saad
  Mar 26 at 1:58
  

  
  
  
  

add a comment | 

$begingroup$

Let $(X,Y) in Bbb X times Bbb Y$ be jointly distributed according
to distribution $P$. Let $h: Bbb X rightarrow tilde Bbb Y$,
where $tilde Bbb Y$ is a predicted output. $ $Let $L(h,P) equiv
Bbb E_P[l(Y, h(X))]$ where $l$ is some loss function.

Show that $f = arg min_h L(h,P) = Bbb E_p[Y mid X = x]$ if $l$ is the
square loss function: $l(Y, h(X)) = (y - h(x))^2$

I figured I show this by showing any other $h$ leads to a larger $L(h,P)$ than $Bbb E_P[Ymid X=x]$.

I start with $$Bbb E_P[(y - Bbb E_p[Ymid X=x])^2] le Bbb E_P[(y - h(x))^2]$$

Then expanding we have:

$$Bbb E_P[y^2-2yBbb E_p[Y|X=x] + Bbb E_P[Ymid X=x]^2] le Bbb E_P[y^2 - 2yh(x) + h(x)^2]$$

And simplifying:

$$-2Bbb E_P[y]Bbb E_p[Ymid X=x] + Bbb E_P[Ymid X=x]^2 le -2Bbb E_P[yh(x)] + Bbb E_P[h(x)^2]$$

But from here I'm a little stuck as to how to continue.

Does anyone have any ideas?

probability machine-learning

share|cite|improve this question

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

$endgroup$

share|cite|improve this question

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

share|cite|improve this question

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

edited Mar 17 at 12:39

Bernard

124k742117

edited Mar 17 at 12:39

Bernard

124k742117

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634

asked Mar 17 at 12:33

Oliver GOliver G

1,2761634