# a simpler version of my original variance puzzle question

Hi: I think that I can simplify my original question a great deal so here's my attempt.

Suppose I have a function $f(t) = t-10$. $t$ denotes time and starts at $t = 0$ and the units of time pass discretely so $t=1$, then $t=2$ etc all the way upto $t = 10$. This function is deterministic so obviously $f(t=10) = 0$.

Now suppose I add the following twist that introduces variation and makes the value of $f(t)$ at each time $t$ random. The twist is the following:

At each new time $t$, a random variable, $\epsilon_{t}$ with mean $\mu = 0$ and variance $\sigma^2 = 1$ is drawn randomly. The value of the random variable $\epsilon_{t}$ is then added to the function $f(t)$ so that the function takes on it's deterministic value + the value of that random variable.

So, at say $t = 1$, if $\epsilon_1 = 0.7$, then this is added to $f(1) = -9.0$ which results in the value $-8.30$. This is the new value of $f(t)$ at time $t=1$. Note now what happens next which is probabbly the trickiest part of the setup. The function continues on from there so that the next value of $f(2)$ at time $t = 2$, BEFORE the next normal random variable, $\epsilon_{2}$, is drawn, is now $2-10 + 0.7 = -7.3$ because the $0.7$ from the previous draw of the random variable $\epsilon_{1}$ remains added in to the function. So, the key point to understand is that random variables drawn at each new time $t$ remain in the previously deterministic function going forward.

Of course, the process is repeated so that, a new random variable is drawn at time $t=2$, so that $f(2)$ will be equal to $-7.3 + \epsilon_{2}$. The process continues along this way with $\epsilon_{t}$ being added in to the function $f(t)$ at each new time $t$ until the process ends at time $t = 10$.

Clearly, if the random variables drawn at each of $t = 1,\ldots 10$ were all zero, then the value of the function at $t = 10$ would be $f(10) = 0$. So, the expected value of $f(10) = 0$. But, the ending value now has variance because of the possibility of the $\epsilon_{t}$ not being $0$.

An added assumption is that, if $f(t)$ happens to cross $0$ at any time $t$, then the process stops at that time $t$ and can be considered to stop exactly at $0$. So, if $f(t)$ reaches the value of $0$ before $t=10$, the end value of $f$ is still considered to be $0$.

Of course, there is also the possibility that, at $t =10$, the process has not crossed $0$ yet. The process would then stop at $t = 10$. So, the process is assumed to end at the earlier of two events: $t = 10$ and crossing through the t axis, whichever comes first.

As stated previously, given the added randomness, the expected value of $f(10)$ is still equal to $0$. My question is: "What is the variance of the final value of $f$ ?".

This seems like a reasonable question but I don't think it's so easy. It may have something to do with brownian motion and stopping times. I really don't know but, even if it is related to that, I have no idea how one would go about solving it. Any references or solutions or thoughts are appreciated. Thanks a lot.

P.S: For those familiar with finance, this problem is sort of related to the notion of "semi-variance" because, in this problem, some of the variance is not actually "bad" variance in that some of it helps the process reach its target earlier than it normally would. Also, thanks to Whuber who gave me the heads up that my original description was too long.

• If this is a restatement of stats.stackexchange.com/questions/364294/…, then please delete the original. – whuber Aug 29 '18 at 13:44
• Are you just trying to say that $f(10) = \epsilon_1 + \epsilon_2 + \cdots + \epsilon_{10}$? – whuber Aug 29 '18 at 14:10
• @Whuber: Hi. Unfortunately, no. One way to look at it is that : $f(10) =$ (deterministic part of $f(10)) + \epsilon_{1} + \epsilon_{2} + \ldots \epsilon_{10}$ but even that's not really true because the function could cross zero before $t=10$ due to the randomness of the $\epsilon_{t}$ coming in at each time $t$. I probably should have called the random function $f^{*}$ and the original deterministic function $f$. – mlofton Aug 29 '18 at 15:35
• @Whuber: I just realized that what you said is close to true because $f(10) = 0$ from a deterministic point of view. But, as I said, once the $\epsilon_{t}$ start being added in AND KEPT IN, one can't really say that since crossing can occur early. Clearly, if what you said was true, then the variance would be $10 \times \sigma^2$. – mlofton Aug 29 '18 at 15:44
• I don't understand what you mean by "crossing." Only you can determine whether my interpretation accords with your question--that's what I'm trying to find out. And if it does, your conclusion that the variance is $10\sigma^2$ would be correct. This is a random walk (on the integers), pure and simple. The function $f$ represents an additive "drift" that affects the expectation but not the variance, and so can be ignored. I hope these thoughts might help you simplify your statement of the question considerably. – whuber Aug 29 '18 at 15:47