# autocorrelation for truncated normal distribution

Mathematics Asked by bearcub on August 17, 2020

I have a (symmetric) truncated normal distribution from which I’m sampling. There is also auto-correlation of the values, so I know:
$$mu$$, $$sigma$$, $$text {autocorr} = a$$.

All three of these values are single-valued. E.g., $$mu_R = 1$$, $$sigma_R = 1$$, $$a_R = 0.5$$.

My understanding of the application of autocorrelation to a series, $$R sim N (mu_R=1,sigma_R=1)$$, with the normal distribution results in the new series $$S$$ as:

$$S(1)=R(1)$$
$$S(i)=a_Rcdot R(i-1)+w(i)$$,
where $$w(i)sim N(0,sigma_w)$$.

(taken from these notes – http://home.cc.umanitoba.ca/~godwinrt/7010/topic8.pdf)

However, I have also seen this formulation:

$$R sim N (mu_R=1,sigma_R=1)$$,
$$b = sqrt{1-a_R^2}$$,
$$S(1)=R(1)$$,
$$S(i)=a_Rcdot S(i-1)+bcdot R(i)$$.

Where does the $$b$$ come from?

Why are we scaling the second term instead of drawing from $$w(i)$$?

Why is the autocorrelation coefficient recursively applied via the first term? (e.g., $$S(2)approx a_Rcdot S(1)$$, $$S(3)approx a_Rcdot S(2)approx a_Rcdotbig(a_Rcdot S(1)big)$$)

The only example I’ve found is this: How to transform/shift the mean and standard deviation of a normal distribution?

But it doesn’t fully explain these questions.

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