# Finding bias of the coin from noise corrupted signals

• I
Summary:
I have been given a dataset of noise corrupted sample with noise having a gaussian PDF. How do I find the bias of the coin in the given problem statement ?
Suppose there are two persons A and B such that both have a personal communication system which can transmit and receive bits. B has a biased coin whose bias is not known. A asks B to toss the coin 2000 times, send a 0 when a tail comes up and a 1 when a head comes up. It is known that whatever A receives is corrupted by noise, which has a Gaussian PDF with mean μ and variance σ2 . A put’s an additional request to B and asks B to simply send 200 zeros before sending the coin toss results. Using these 2200 samples of data, find the mean, variance of noise and also the bias of the coin.

My attempt:

From the 200 zeros that are sent first, we can determine the noise parameters like mean and variance because N + 0 = N , where N is the noise. But how do I find the bias of the coin from the remaining 2000 samples ?

Can anyone help me with the right approach to this problem ?

mathman
The effect of the noise is unknown. since the test did not show what happens when 200 1's are sent. My guess use the mean of the 2000 flips to determine bias.

• Amitkumarr
The effect of the noise is unknown. since the test did not show what happens when 200 1's are sent. My guess use the mean of the 2000 flips to determine bias.
How do I determine bias from the mean of the 2000 flips ? I know that we can take Y=N+S where Y is the signal received at B including the noise(N) and S is the original signal sent by A. Noise is assumed to be additive and I know mean and variance of N and Y. Then how should I proceed ?

mathman

• Amitkumarr
WWGD
Gold Member
How do I determine bias from the mean of the 2000 flips ? I know that we can take Y=N+S where Y is the signal received at B including the noise(N) and S is the original signal sent by A. Noise is assumed to be additive and I know mean and variance of N and Y. Then how should I proceed ?
I think the bias would be the mean minus 1000, which is the number of heads/tails in a fair coin.

I think the bias would be the mean minus 1000, which is the number of heads/tails in a fair coin.
Thanks for the reply. I have one small doubt:- If Z=X + Y where X is a continuous random variable and Y is a discrete random variable then is E[Z]=E[X] + E[Y] true for this case ? where E[Z] is the expectation value of Z.

WWGD
Gold Member
Thanks for the reply. I have one small doubt:- If Z=X + Y where X is a continuous random variable and Y is a discrete random variable then is E[Z]=E[X] + E[Y] true for this case ? where E[Z] is the expectation value of Z.
Yes, expectation is always linear( as long as individual expectations exist, of course): Expectation of sum is the sum of the expectations.

• Amitkumarr
Yes, expectation is always linear( as long as individual expectations exist, of course): Expectation of sum is the sum of the expectations.
Thanks, now I have got an idea of how to solve this problem.

• WWGD
WWGD
Gold Member
Thanks, now I have got an idea of how to solve this problem.
Just curious, are you using H/N=H/2000 : Number of heads ( equiv., tails) as an estimator?

It appears to me that you are trying to find the bias of the noise. You can't know that from what you are given. I think they are assuming the noise in unbiased. It is safe for you to assume that too because otherwise you can't get the answer. If you really want to protect yourself you can make a note of this in your answer.

Usually in statistics you can't get a pure sample of the noise, so seldom could you measure the bias of the noise. Since you don't know what it is, you just assume it is zero. It's the best you can do.

Stephen Tashi