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I have a random variable X, to which additive gaussian noise is added resulting in random variable Y.

I have $\log \frac{X=-1|Y}{X=1|Y}$. This is equal to $\log \frac{P\left\{X=-1\right\} \cdot P\left\{Y|X=-1\right\} \cdot \frac{1}{p(y)}}{P\left\{X=1\right\} \cdot P\left\{Y|X=1\right\} \cdot \frac{1}{p(y)}}$ which reduces to $\log \frac{P\left\{X=-1\right\}}{P\left\{X=1\right\}}$ + $\log \frac{P\left\{Y|X=-1\right\}}{P\left\{Y|X=1\right\}}$. The second term in the addition can now be represented as $\log \frac{\frac{1}{\sqrt{2\cdot\pi\cdot\sigma^2}} \cdot e^{-\frac{(y-(-1))^2}{2\cdot\sigma^2}}}{\frac{1}{\sqrt{2\cdot\pi\cdot\sigma^2}} \cdot e^{-\frac{(y-(1))^2}{2\cdot\sigma^2}}}$ which can be simplified after cancellation to $\log e^{\frac{(y+1)^2 - (y-1)^2}{2\sigma^2}}$ which again reduces to $\frac{2\cdot y}{\sigma^2}$

This was possible because we used the Gaussian distribution for noise. Can we do something similar while computing $\log \frac{P\left\{Y|X=-1\right\}}{P\left\{Y|X=1\right\}}$ given an i.i.d. random variable and its probability mass function?

Rohit
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  • Sorry but what exactly is i.i.d. here? Do you know what i.i.d. means? – Did Dec 08 '13 at 12:34
  • i.i.d. is independent and identically distributed random variables.

    For e.g: I have n = (-3,-2,-1,0,1,2,3) with p(n) = (0.1,0.1,0.2,0.2,0.2,0.1,0.1) respectively. This has mean = 0 and variance = 3.

     – Rohit Dec 08 '13 at 12:35
  • Suspicion confirmed, you do not know what i.i.d. means. A single random variable is neither i.i.d. nor not i.i.d., the phrase is just absurd. – Did Dec 08 '13 at 13:26
  • What about a sequence of fair coin flips? Okay to be precise, I have a sequence of noise samples where $n_k \in \left{-3,-2,-1,0,1,2,3\right}$ – Rohit Dec 08 '13 at 18:21
  • What about it? Tell me. – Did Dec 08 '13 at 18:28

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