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Old 01-16-2013, 11:41 AM
ripande ripande is offline
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Default Query regarding Noise Targets lecture 4

As per lecture 4 :- Slide 16/22

Noise target = Diterministic Target + Noise (This is clear)

However it then says :

Diterministic target f(x) = E(Y|X). Hence
y ( Noisy target) = E(Y|X) ( Diterministic target) + noise ( y-f(x) )

I do not understand the above statement. Did we not say that to accomodate for noisy targets we replace y = f(x) with a conditional distribution of y given x. Should then not the noisy target be defined by
E(Y|X) ?
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Old 01-16-2013, 11:59 AM
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yaser yaser is offline
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Default Re: Query regarding Noise Targets lecture 4

Quote:
Originally Posted by ripande View Post
As per lecture 4 :- Slide 16/22

Noise target = Diterministic Target + Noise (This is clear)

However it then says :

Diterministic target f(x) = E(Y|X). Hence
y ( Noisy target) = E(Y|X) ( Diterministic target) + noise ( y-f(x) )

I do not understand the above statement. Did we not say that to accomodate for noisy targets we replace y = f(x) with a conditional distribution of y given x. Should then not the noisy target be defined by
E(Y|X) ?
The noisy target is y and indeed it is specified by P(y|{\bf x}). The equation decomposes the noisy target into a noiseless component plus a pure noise component.
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Old 01-16-2013, 12:08 PM
ripande ripande is offline
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Default Re: Query regarding Noise Targets lecture 4

Is E(Y|X ) the noiseless component ? If yes, how ?
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Old 01-16-2013, 12:11 PM
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yaser yaser is offline
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Default Re: Query regarding Noise Targets lecture 4

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Originally Posted by ripande View Post
Is E(Y|X ) the noiseless component ? If yes, how?
Yes, it is. It is a deterministic function of {\bf x}. The noisy aspect has been integrated out (averaged out) by the expected value.
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Old 01-16-2013, 12:24 PM
ripande ripande is offline
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Default Re: Query regarding Noise Targets lecture 4

Understood. Thanks prof Yaser
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Old 04-13-2013, 10:45 AM
jlaurentum jlaurentum is offline
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Default Re: Query regarding Noise Targets lecture 4

One question:

If X is a random sample, it is clearly a random variable (or rather, a sequence of random variables).

If Y is also a random variable (the target), then wouldn't E(Y|X) be a random variable and not a deterministic quantity? (unless X and Y are independent variables in which case the conditional expectation collapses to a constant?)
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Old 04-13-2013, 11:20 AM
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yaser yaser is offline
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Default Re: Query regarding Noise Targets lecture 4

Quote:
Originally Posted by jlaurentum View Post
One question:

If X is a random sample, it is clearly a random variable (or rather, a sequence of random variables).

If Y is also a random variable (the target), then wouldn't E(Y|X) be a random variable and not a deterministic quantity? (unless X and Y are independent variables in which case the conditional expectation collapses to a constant?)
Let me first use the class notation to make sure we are talking about the same thing. The data (sample) is ({\bf x}_1,y_1), \cdots, ({\bf x}_N,y_N) where each {\bf x}_n is (often) a random vector (a vector of random variables) and each y_n is a random variable. The probability distribution P(y|{\bf x}) is the conditional distribution of that random variable given {\bf x}.

The conditional expectation, {\bf E}(y|{\bf x}), is a deterministic quantity (more presisely, a deterministic function of {\bf x}). As you point out, if {\bf x} and y are statistically independent, that function is a constant (independent of {\bf x}). However, even if {\bf x} and y are not statistically independent, that function is still a deterministic, albeit non-constant, function of {\bf x}.
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Old 04-13-2013, 01:02 PM
jlaurentum jlaurentum is offline
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Default Re: Query regarding Noise Targets lecture 4

Oh ok. I get it, Professor. It's like with the regression, where for a given linear model: Y=\beta_0+\beta_1\cdot X_1 + \ldots, what we're actually doing is specifying E(Y|X)=\beta_0+\beta_1\cdot X_1 + \ldots.
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