Autocorrelation of a Bernoulli Coin Flipping Experiment

[tworitdash]

I am confused at one point. The coin flipping Bernoulli Process has a probability of p of getting HEADS and a probability of 1-p of getting TAILS. Let's define a random variable x[n], which takes the value +1 when it is a HEADS, and -1 when it is a TAILS. The mean or estimation of x[n] becomes (2p -1) and I can derive it as an integration of probability function with the function itself. Which results in the sum of +1 times p and -1 time (1-p). However, when it comes to autocorrelation if the lag is 0, That is Estimate[x[n+m]x[n]] = 1 when m = 0 and if m is not equal to 0, it becomes (2p -1) ^ 2. I basically get it when I try to understand the physical meaning of it, but mathematically how this calculation is done? Because of the fact that the autocorrelation is a second-order property, it becomes (2p - 1) ^ 2. However, not for m = 0. Why and how?E{x[n]} = 2p−1

E{x[n+m]x[n]} = 1 for [m = 0]
E{x[n+m]x[n]} = (2p - 1)² for [m != 0]

 

[RPinPA]

If p = 0.5, then 2p - 1 = 0. That means that a flip is correlated with itself ($m=0$) but any two different flips ($m\neq 0$) are completely uncorrelated. That's what you'd expect. What is confusing you about that?

 

[vela]

That is Estimate[x[n+m]x[n]] = 1 when m = 0 and if m is not equal to 0, it becomes (2p -1) ^ 2. I basically get it when I try to understand the physical meaning of it, but mathematically how this calculation is done? Because of the fact that the autocorrelation is a second-order property, it becomes (2p - 1) ^ 2. However, not for m = 0. Why and how?

Presumably, you calculated the probabilities that x[n]=1 and x[n+m]=1; x[n]=1 and x[n+m]=-1; x[n]=-1 and x[n+m]=1; and x[n]=-1 and x[n+m]=-1. Are those probabilities the same when $m=0$ and $m\ne 0$?

 

[tworitdash]

Presumably, you calculated the probabilities that x[n]=1 and x[n+m]=1; x[n]=1 and x[n+m]=-1; x[n]=-1 and x[n+m]=1; and x[n]=-1 and x[n+m]=-1. Are those probabilities the same when $m=0$ and $m\ne 0$?

Yes It should be the same when m = 0. Isn't it? But is it an expession of p or just 1?

 

[Ray Vickson]

I am confused at one point. The coin flipping Bernoulli Process has a probability of p of getting HEADS and a probability of 1-p of getting TAILS. Let's define a random variable x[n], which takes the value +1 when it is a HEADS, and -1 when it is a TAILS. The mean or estimation of x[n] becomes (2p -1) and I can derive it as an integration of probability function with the function itself. Which results in the sum of +1 times p and -1 time (1-p). However, when it comes to autocorrelation if the lag is 0, That is Estimate[x[n+m]x[n]] = 1 when m = 0 and if m is not equal to 0, it becomes (2p -1) ^ 2. I basically get it when I try to understand the physical meaning of it, but mathematically how this calculation is done? Because of the fact that the autocorrelation is a second-order property, it becomes (2p - 1) ^ 2. However, not for m = 0. Why and how?E{x[n]} = 2p−1

E{x[n+m]x[n]} = 1 for [m = 0]
E{x[n+m]x[n]} = (2p - 1)² for [m != 0]

If the successive flips are "independent", the $X[n]$ results are uncorrelated; that is, the auto-covariance is
$$\text{Cov}(X[n], X[k]) = 0 \; \text{for} \; n \neq k.$$ Since
$$\text{Cor}(X[n],X[k]) \equiv \frac{\text{Cov}(X[n],X[k])}{\sigma_{X[n]} \sigma_{X[k]}}, $$
(where $\sigma_X$ is the standard deviation of $X$) it follows that the correlation is zero as well.
Remember: the covariance $\text{Cov}$ is defined as
$$\text{Cov}(X[n],X[k]) \equiv E[ (X[n]-E X[n]) (X[k] - E X[k])], $$
and this evaluates to
$$\text{Cov}(X[n],X[k]) = E (X[n] X[k] ) - (E X[n]) (E X[k]).$$ This is zero for a Bernoulli process.For $n = k$ the covariance reduces to the variance:
$$ \text{Var} (X[n]) = E( X[n]- E X[n])^2 = E(X^2[n]) - (E X[n])^2 = 4p(1-p).$$