How to model this distribution using Central Limit theorem?

I have distribution that can be defined as below,

$S=a_0\cdot b_0 + a_1\cdot b_1 + a_2\cdot b_2 + \cdots +a_{n-1}\cdot b_{n-1}$

Now, I want find the distribution of $S$ when, $a_i$'s are selected from a certain distribution with standard deviation $\sigma$ (for simplicity we can assume it a Gaussian distribution). And $b_i$'s can be $+5$ with probability $p$ and $-5$ with probability $1-p$. How, as far as I know $S$ will be distributed normally too, but what will be the standard deviation of such distribution.

• The variance of the limiting distribution after scalling by $\sqrt{n}$ is $Var(a_0b_0)$. It can be computed if $a_0$ and $b_0$ are independent. – Sayan May 26 '17 at 14:43
• @sayan I did not get your answer. Can you illustrate a little more? Again for simplicity we can take the value $p=1/2$ – Rick May 26 '17 at 14:46
• If $(a_i)$ is i.i.d. centered normal with variance $\sigma^2$, if $(b_i)$ is i.i.d. with $P(b_i=+5)=p$, $P(b_i=-5)=1-p$, and if $(a_i)$ and $(b_i)$ are independent, then $S$ is centered normal with variance $25n\sigma^2$ (and standard deviation $5\sigma\sqrt{n}$). If $(a_i)$ is not centered, no chance for such a nice result. – Did May 26 '17 at 15:50
• whoever downvoted can please provide the reason? – Rick May 29 '17 at 10:01

$S$ will not be normally distributed, but no matter

If everything is independent then

• $a_i$ has mean $\mu$ and variance $\sigma^2$ so $E[a_i^2]=\sigma^2+\mu^2$
• $b_i$ has mean $5-10p$ and $E[b_i^2]=25$
• $a_i b_i$ has mean $\mu(5-10p)$ and $E[a_i^2 b_i^2]=E[a_i^2]E[b_i^2]= 25(\sigma^2+\mu^2)$ so has variance $25\sigma^2 +100p \mu^2-100p^2 \mu^2$
• $S=\sum a_i b_i$ has mean $n\mu(5-10p)$ and variance $25n(\sigma^2 +4p(1-p)\mu^2)$ and so a standard deviation which is the square-root of that
• if $p=\frac12$ then the mean of $S$ is $0$ and the standard deviation of $S$ is $5\sqrt{n(\sigma^2+\mu^2)}$
• If $\mu=0$ (plus some other conditions), then S is normal, see my comment on main. – Did May 26 '17 at 15:52
• @Did Yes - though $\mu=0$ is not in the question. In that special case, the distribution of $a_ib_i$ is the same as the distribution of $5a_i$ and so $p$ becomes irrelevant – Henry May 26 '17 at 16:42
• @Henry Thank you for kind help. Will this change in any way if the distribution of $a$ is discrete instead of continuous? (we can take $\mu\ =\ 0$). We can think of a man in the middle of the plank of length $2t$. The man takes $n$ steps. At each step, he samples a value $x \leftarrow D_\sigma$ and either takes $5x$ step forward or $5x$ step backward with probability $p\ =\ 1/2$. How big should the plank be such that the man falls of from the plank with a very low probability? – Rick May 27 '17 at 8:13
• @Ishan - my calculations of the mean, variance and standard deviation require independence of everything, but nothing else. So if $\mu=0$ the mean of $S$ will be $0$, the variance will be $25n\sigma^2$ and the standard deviation $5\sqrt{n}\sigma$, for any $p$. If the $a_i$s are discrete random variables then $S$ will be too – Henry May 27 '17 at 8:21