# A tricky question about sequences of Normal RVs

Suppose I project a sum of money that will be normally distributed (e.g., tomorrow I will have some amount that is drawn from a normal population, where the mean is 100 dollars and the s.d. is 10%, or 10 dollars.) Now, I put it in the bank. The next year I will earn interest on it, which itself is normally distributed with mean 10% and s.d. 10% . I am trying to predict the RV for then - after the year.

The tempting thing is to look at the RV for tomorrow, which is $N(100,10)$ and say that the return has the shape $N(10%,10%)=N(10,10)$ since 10 is 10% of the expected value of 100 dollars.

But it strikes me the return RV is, to some degree, correlated with the initial amount. For example, it's obvious after I receive the first bit that if I only get 85 dollars, then the RV for the return over the next year is distributed as $N(8.5, 8.5)$.

What would the total RV actually be for the end of that second period, as calculated now? I am simply adding two normally distributed RVs. At times I can convince myself the two are uncorrelated; the return RV is simply being scaled by the result of the first. At other times I convince myself the correlation is 1.0,since the return RV is a perfect slice (if you will) of the result of period 1. But it also has a random element, so that's not quite right either.

Another way to look at this is this: if the bank didn't look at my balance, but put put in two random amounts that were normally distributed with known means and s.d.s, then it would surely be just the sum of two independent RVs. But, the random amount the bank will add will, in fact, be based on the outcome of the first experiment. So what is the nature, and how to a calculate, the expected distribution now considering the two back-to-back process that are interlinked. My intuition tells me it is normal, but with some correlation.

$A\sim N(100, 10)$: amount tomorrow
$R\sim N(0.1, 0.1)$: yearly discrete return
$B=A\cdot(1+R)$: amount after one year

$$E[B] = E[A]+E[AR] = E[A]+cov(A,R)+E[A]E[R] \\ \underbrace{=}_\text{if A and R are independant} E[A]+E[A]E[R]=110 \\ var(B) = var(A+AR) = var(A)+var(AR) + 2cov(A, AR)\\ = var(A) + E[A^2 R^2] - E[AR]^2 + 2(E[A^2R]-E[A]E[AR]) \\ = var(A) + E[A^2] E[R^2] - (E[A]E[R])^2 + 2(E[A^2]E[R]-E[A]E[A]E[R]]) \\ = var(A) + (var(A)+E[A]^2)(var(R)+E[R]^2) - (E[A]E[R])^2 + 2((var(A)+E[A]^2)E[R]-E[A]E[A]E[R]]) \\ = 100+(100+100^2)\cdot((0.01+0.1^2))- 100^2\cdot 0.1^2+ 2\cdot((100+100^2)\cdot 0.1-100\cdot 100\cdot 0.1) = 222$$

Checking the results using a Monte-Carlo simulation on R:

iter <- 1e7
A <- rnorm(iter)*10+100
R <- rnorm(iter)*0.1+0.1
B <- A*(1+R)
print(mean(B))
> 110.0011
print(var(B))
> 221.9952


Sorry for the mess, feel free to ask if you need clarifications about any of the steps. Also if you're interested in this kind of problems, you should look into Brownian motions. They're widely used in finance to model price series.

• That is a genius way to look at it. It also explains a simple Monte Carlo I was doing that gave a puzzling outcome, but I think I see why, roughly (if I actually do a number of numeric trials, then look at the correlation, it is amazingly insensitive to $\sigma$, but leads to a 5%-10% 'error' in std deviation depending on $\mu$, in line with your simulation). I have been looking into Brownian motions, but it is near impossible to find anyone or text who can explain and reconcile what it all means. How did you get to 4th line on variance, where $2(E[A2]E[R]−E[A]E[A]E[R]])$ 'disappeared'? Commented May 23, 2018 at 22:19
• It didn't disappear, it's on the next line :) $$2(E[A^2]E[R]-E[A]E[A]E[R]]) = 2((var(A)+E[A]^2)E[R]-E[A]E[A]E[R])$$ Commented May 23, 2018 at 22:32
• There's a good explanation of Brownian motion in week 8 of this free course: coursera.org/learn/financial-engineering-1 Commented May 23, 2018 at 22:42

First of all, rather than looking at the interest, it's better to look at the factor it's being multiplied by. After one year, the amount you have is:

$B_1 = B_0 \cdot F_1$

where $F_1$ is normally distributed with mean 1.1 and sd 0.1

The amount that you have after two years is

$B_2 = B_1 \cdot F_2$

$F_1$ and $F_2$ are identically distributed, both mean 1.1 and sd 0.1 . And now, with this formulation, they are completely independent. We can also write it as

$B_2 = B_0 \cdot F_1\cdot F_2$

or, if we want to express it as a sum rather than product,

$log(B_2) = log(B_0)+log( F_1)+log( F_2)$

Note that modelling the interest as being normally distributed is problematic in that if the interest is less than -100%, then that would mean you end up with a negative amount. Here, that would mean a z score less than -10, so the probability is negligible, but from a theoretical standpoint, it's a problem.

• Yes, I am trying to do an example I can explain to others whose math is not strong. One challenge with the approach I think you are using is that, if I am not mistaken, the results become non-normal, hence the urge to add normals with the risk there is correlation. Essentially, I think you are saying, 'go log-normal' as one way out. Commented May 23, 2018 at 22:28