$$X \sim N(\mu_1,\sigma_1^2)$$ $$Y \sim N(\mu_2,\sigma_2^2)$$ 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 $Var(\sum_1^nX_t)=n\cdot1$

  • 1
    $\begingroup$ Hint: If $$Y_j = \sum_{i=1}^j X_i$$ where $X_i \sim {\rm Normal}(0,1)$, are IID, then are the $Y_j$'s independent? Are they even identically distributed? $\endgroup$
    – heropup
    Commented Mar 7, 2014 at 23:21
  • $\begingroup$ The expected value of the mean is $0$, but if you only look at one case then the sample mean will probably not be. $\endgroup$
    – Henry
    Commented Mar 8, 2014 at 0:15
  • $\begingroup$ What about the variance?? What should the variance then be of iterativly summing the variables? $\endgroup$
    – jessica
    Commented Mar 8, 2014 at 0:30
  • $\begingroup$ Please help!! :) $\endgroup$
    – jessica
    Commented Mar 8, 2014 at 0:32
  • $\begingroup$ You're on your way to defining a Brownian motion process! ;=) $\endgroup$
    – JPi
    Commented Mar 8, 2014 at 1:37

1 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.

  • $\begingroup$ H Henry, thank you for responding. Where does the $(n+1-i)$ come from when your computing the mean? $\endgroup$
    – jessica
    Commented Mar 8, 2014 at 0:57
  • $\begingroup$ @jessica: $X_1$ appears $n$ times in the calculation, $X_2$ appears $n-1$ times, ..., $X_n$ appears $1$ time. $\endgroup$
    – Henry
    Commented Mar 8, 2014 at 1:22

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