- #1
six7th
- 14
- 0
Suppose I have a Gaussian probability distribution:
[itex]N_{A}(0,1).[/itex]
A set of values are generated from this distribution to which an arbitrary amount of Gaussian noise, say [itex]N_{B}(0,0.5)[/itex], is added and then the [itex]N_{B}[/itex] values sorted from lowest to highest. These are then digitised by assigning 0 to a value below 0 and then 1 to a value above 0. We end up with values such as:
-3.11359 | 0 | -6.91717 | 0 |
-3.31609 | 0 | -6.70810 | 0 |
-3.67497 | 0 | -6.13978 | 0 |
-2.73024 | 0 | -6.12173 | 0 |
-4.20178 | 0 | -6.07266 | 0 |
-3.38846 | 0 | -5.88277 | 0 |
To calculate the mutual information in bits between this set of data we use the following formula:
[itex]I(A;B) = \sum_{x \in A} \sum_{y \in B}P(x,y)log_{2}(\frac{P(x,y)}{P(x)P(y)}) [/itex]
where [itex]P(x,y)[/itex] is the probability of A = x and B = y. Now in this case P(0,1) = P(1,0) = P(1,1) = 0 so we are left with
[itex]I(A;B) = P(0,0)log_{2}(\frac{P(0,0)}{P(x=0)P(y=0)})[/itex]
Now here is what I don't get. For this Gaussian distribution P(x=0) ≈ 0.5 and P(y=0) ≈ 0.5 and from this set of data P(0,0) = 1. This gives,
[itex]I(A;B) = log_{2}(\frac{1}{0.25}) = 2 [/itex]
From what I understand it is should not possible for the mutual information to be greater than 1. Where am I going wrong with this? I feel as though I am making a very basic error.
Thanks.
[itex]N_{A}(0,1).[/itex]
A set of values are generated from this distribution to which an arbitrary amount of Gaussian noise, say [itex]N_{B}(0,0.5)[/itex], is added and then the [itex]N_{B}[/itex] values sorted from lowest to highest. These are then digitised by assigning 0 to a value below 0 and then 1 to a value above 0. We end up with values such as:
-3.11359 | 0 | -6.91717 | 0 |
-3.31609 | 0 | -6.70810 | 0 |
-3.67497 | 0 | -6.13978 | 0 |
-2.73024 | 0 | -6.12173 | 0 |
-4.20178 | 0 | -6.07266 | 0 |
-3.38846 | 0 | -5.88277 | 0 |
To calculate the mutual information in bits between this set of data we use the following formula:
[itex]I(A;B) = \sum_{x \in A} \sum_{y \in B}P(x,y)log_{2}(\frac{P(x,y)}{P(x)P(y)}) [/itex]
where [itex]P(x,y)[/itex] is the probability of A = x and B = y. Now in this case P(0,1) = P(1,0) = P(1,1) = 0 so we are left with
[itex]I(A;B) = P(0,0)log_{2}(\frac{P(0,0)}{P(x=0)P(y=0)})[/itex]
Now here is what I don't get. For this Gaussian distribution P(x=0) ≈ 0.5 and P(y=0) ≈ 0.5 and from this set of data P(0,0) = 1. This gives,
[itex]I(A;B) = log_{2}(\frac{1}{0.25}) = 2 [/itex]
From what I understand it is should not possible for the mutual information to be greater than 1. Where am I going wrong with this? I feel as though I am making a very basic error.
Thanks.