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I generate $N$ independent uncorrelated random numbers, normally distributed with mean $\mu$ and variance $\sigma^2$ and place them into an array $a[i]; 1 \le i \le N$.

I then compute the sum $A = \sum_{i=1}^{N} a[i]$ of the array. Then $A \sim N(N \mu, N \sigma^2)$.

Now I copy the array $a$ to another array $t$ and select $k$ elements in $t$ at random, $1 < k < N$, and reverse the sign of those elements.

$t[i] = -a[i]$ if $i$ is one of the $k$ randomly selected elements.

$t[i] = a[i]$ otherwise.

Then I compute the sum $T = \sum_{i=1}^{N} t[i]$.

Fixing the original array $a$ and the number of elements $k$ to negate, I repeat many times the copying to $t$, each time selecting a new random $k$ elements to negate and summing to get a new $T$. My question is, what is the mean and variance of the series of $T$ values?

Numerical trials suggest the mean of $T$ is $N \mu (1 - 2k/N)$ which seems reasonable. What I'm hoping for is a formula for the variance of $T$ in terms of $A, k, N,\mu$ and $\sigma$.

The problem arises in my attempts to calculate the error rate of a binary linear block code. $N$ is the length of a codeword, $a[]$ is a received noisy all-zero codeword with $A$ as its log likelihood, and $k$ is the hamming weight of codewords which might be mistaken for the all-zero message if $T$ is large enough.

Ultimately what I need is P(T > A | a[], k) which is the probability that a codeword of weight $k$ will exceed the likelihood of the correct message (which has weight zero). I then have to integrate over the probability distribution of A and over the weight spectrum of my block code.

If I just assume $\operatorname{Var}(T) = \operatorname{Var}(A)$, I get a fairly weak upper bound on the code performance. For example predicting 74% error rate when the code actually achieves 57%. I seem to need roughly $\operatorname{Var}(T) \approx 0.9 \operatorname{Var}(A)$ to get good predictions when integrating over the weight spectrum.

I ran some numerical experiments. Each experiment begins by filling the array $a[]$ with random samples from $N(\mu,\sigma^2)$ and calculating the sum $A$. The experiment then makes $10^8$ trials in which $a[]$ is copied to $t[]$ and a randomly selected $k$ elements are negated in $t[]$. The experiment ends by logging $A$ and the mean and variance of the $10^8$ $T$ values. Fixing $N=352, \mu=1, \sigma=3.2$ the experiments span a range of $k$.

For $k=48$ a scatter plot of $\operatorname{Var}(T)$ against $A$, 160102k48.gif with one point per experiment. There seems to be no correlation with $A$ which simplifies things. For $k=48$ the average of $\operatorname{Var}(T)$ is 1706. ($\operatorname{Var}(A) = 3604.48$). Plotting the average of $\operatorname{Var}(T)$ for a range of $k$ gives 160102kp.gif. This nice curve is what I'm looking for a formula for. It seems that once $k \ge N/2$ the $\operatorname{Var}(T)$ is practically equal to $\operatorname{Var}(A)$. This is not so important, it is the lower values of $k$ (lower weight codewords) which are the more significant for code performance.

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  • $\begingroup$ I tried to improve TeX code. However, do you mean $1 \le i \le N$ at the start and then $1 < k < N$ for the elements selected? $\endgroup$ – BruceET Jan 1 '16 at 23:35
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Due to the observations being IID, it makes no difference which $k$ of the $N$ observations in the sample we negate. It suffices to assume that the first $k$ observations are negated. Then these first $k$ observations are each normally distributed with mean $-\mu$ and variance $\sigma^2$. Their sum is normal with mean $-k\mu$ and variance $k\sigma^2$.

Similarly, the remaining $N - k$ observations not negated have a sum that is normal with mean $(N-k)\mu$ and variance $(N-k)\sigma^2$.

Therefore, the resulting total sum is normal with mean $$-k\mu + (N-k)\mu = (N - 2k)\mu,$$ and variance equal to the sum of their component variances; i.e., $$\operatorname{Var}[T] = k\sigma^2 + (N-k)\sigma^2 = N\sigma^2.$$ This last property arises because although the sum of the first $k$ observations and the sum of the last $N-k$ observations are not identically distributed, they remain independent; therefore, the variance of their sum is equal to the sum of their variances. Throughout, we exploit the fact that the sum of independent normal variates is itself normal.

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  • $\begingroup$ Note that I am filling $a[]$ just once from my iid source of noisy samples, then making observations of $T$ while $A, a[]$ and $k$ are fixed. If every trial selects the first $k$ to negate I will always get the same $T$ value. $\endgroup$ – Paul Jan 2 '16 at 5:04
  • $\begingroup$ @Paul Based on my understanding of the conditional probability you wish to calculate, I am not sure if the requested parameters for T will be sufficient for you to actually calculate what you want. The reason is that the conditional distribution of $T$ given the sample is not normally distributed. To see why, suppose $N = 2$ and the sample observed is $\boldsymbol a = (0,1)$. Then $T \mid \boldsymbol a, k$ isn't even continuous let alone normal. $\endgroup$ – heropup Jan 2 '16 at 17:17

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