Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have a relationship as follows:

$$Y = a X + G(0,\sigma^2),\text{ so }y = a X + \text{some Gaussian noise}.$$

The conditional probability distribution of $y$ given $x$, i.e. $P(y|x)$, is equal to a Gaussian with mean $= a X$ and variance $= \sigma^2$.

I intuitively understand this as the expected value for $y$ should be $a X$ and this will vary due to the noise with the same variance of the noise. Is there a formal proof for this?

Thanks, Aly

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Instead of a 'formal' proof, I suggest you be sure to understand these simple facts:

  1. If $Z = k + Y$, where $k$ is a constant, then the probability density of $Z$ is the same of $Y$ except for a shift : $f_Z(z) = f_Y(z-k)$. As corolary: $E(Z) = k + E(Y)$ and $Var(Z) = Var(Y)$.

  2. If $Z = X + Y$ and we condition on $X$ (which means that we are given the value of $X$), then $X$ can be regarded as a constant, and the above applies. More formally (but, I insist, this should not be necessary to understand the previous) $f_{Z|X}(z|x) = f_Y(z-x)$

share|improve this answer
    
I think you should write $f_Z(z)=f_Y(z-k)$ rather than $f_Z(Z)-f_Y(Z-k)$, i.e. don't use the capital $Z$ to refer to both the random variable and the argument to the density function. –  Michael Hardy Apr 11 '12 at 21:06
    
@MichaelHardy: you're right, fixed. –  leonbloy Apr 12 '12 at 0:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.