Sum of Random Distributions/ Unusual Results

Question

$$X \sim N(\mu_1,\sigma_1^2)$$ $$Y \sim N(\mu_2,\sigma_2^2)$$ then $$X+Y \sim N(0,\sigma_1^2+\sigma_2^2)$$

One way, I tested this to be true is in excel, I used the norm.inv(rand(),0,1) and created an array of 1000 rows/data points.

          X            Y
1   -0.57306826      0.516810296
2   -0.209113627     0.191298912
3   -1.399749083    -1.195672984
4   1.317783869     0.003841951
5   1.800761285     0.866364269
6   1.259689933     -0.985409706
7   -0.501198314    1.799725917
8   0.209555354     -0.258582777
9   -0.744123211    0.738373998
10  0.595194985     -0.653501771

Then I summed $X$ and $Y$ and then took the average of the two columns and I indeed got a mean of 0, ($\mu_{x+y}=0$) and a standard deviation ($\sigma_{x+y}=2$).

So I said, perfect!! But then an idea occurred to me. What if started from an initial standard normal value, and then summed another and added it to the former as such, in other words, reiteravily adding normal values.

$X_t=X_{t-1}+X_{t-2}$ , where each $X \sim N(0,1)$

In excel format,

1   0
2   =NORM.INV(RAND(),0,1)+A1
3   =NORM.INV(RAND(),0,1)+A2
4   =NORM.INV(RAND(),0,1)+A3
5   =NORM.INV(RAND(),0,1)+A4
6   =NORM.INV(RAND(),0,1)+A5
7   =NORM.INV(RAND(),0,1)+A6
8   =NORM.INV(RAND(),0,1)+A7
9   =NORM.INV(RAND(),0,1)+A8
10  =NORM.INV(RAND(),0,1)+A9

Using this approach, when I averaged the entire column of 1000 data points, my mean wasn't zero and my variance also wasn't $1000$ as I had expected. What gives? The variance never equaled 1000 throughout all the simulations of random numbers. Theoretically, my $E(\sum_1^nX_t)=0$ and the Variance $Var(\sum_1^nX_t)=n\cdot1$

Answer 1

As heropup says, $Y_j = \sum_{i=1}^j X_i$.

You then take the mean of these $M=\frac1n \sum_{j=1}^n Y_j$ which is $M=\frac1n \sum_{i=1}^n (n+1-i)X_j$.

$M$ is a random variable with expected value $0$ and variance $\frac1{n^2} \sum_{i=1}^n (n+1-i)^2 = \frac{(n+1)(2n+1)}{6n}$ which for $n=1000$ is about $333.8$, so sample values of $M$ up to around $\pm 35$ will not be uncommon.

You also took the sample variance $V=\frac1{n-1} \sum_{j=1}^n \left(Y_j - M \right)^2$ which I cannot be bothered to calculate but empirically is about $\frac{n+1}{6}$ and thus about $166.8$ when $n=1000$, and with a large variance of the variance: the distribution of the variance is right skewed.