# iterative transform of standard normal random variable

Given a discrete series of random variable $n(i)$ that each element follows the standard normal distribution $N(0,1)$, another series is defined iteratively as: $$u(i+1)=au(i)+bn(i)$$ where $0<a<1$ and $b$ any real number.

How to show that $u$ is also a normal-distributed random variable?

What is the variance of $u$ in terms of $a$ and $b$?

Thanks.

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A linear combination of two jointly normal random variables is normal. Note that simply asserting that the $n(i)$ are standard normal random variables does not mean that the $n(i)$ are jointly normal. So you need to look at the $n(i)$ some more. Did it say anywhere (maybe at the beginning of the chapter) that the $n(i)$ are independent standard normal random variables? Independent normal random variables are jointly normal random variables .... – Dilip Sarwate Jul 16 '13 at 3:12
@DilipSarwate Yes you are right. $n(i)$ are independent standard normal random variables and $u(0)=0$. How about the variance of $u$? – MiniUFO Jul 16 '13 at 6:32
So, what is $u(1)$? Can you figure out its mean and variance? If you can, write an expression for $u(2)$ and try to figure out the mean and variance of that. – Dilip Sarwate Jul 16 '13 at 11:39
@DilipSarwate Thanks, I can get the variance of each $u(i)$. Then how about the variance of the whole series $u$? – MiniUFO Jul 17 '13 at 3:46
What is your definition of the variance of a series? As I understand it, for a stationary series, the variance is a constant. For a nonstationary series, the variance is itself a series which you say you have calculated. – Dilip Sarwate Jul 17 '13 at 10:33