# Normal Distribution from Standard Deviation?

So I have a data set $(x_{1},y_{1}), (x_{2},y_{2}),\dots,(x_{n},y_{n})$ and from it I have the values of $\sum x$, $\sum x^{2}$, $\sum y$, $\sum y^{2}$, $\sum xy$.

My question is, how do I find a normal distribution that best fits this data set and how do I use these values to calculate the standard deviation for the normal distribution?

Basically, given a data set, how do I find the values of the mean and standard deviation for the normal distribution of best fit? Are they the same as the mean of the data set?

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It seems that you have bivariate data (i.e. data for a scatterplot). Do you want to approximate the xi or the yi or do you want a bivariate normal distribution? –  Hans Engler Dec 22 '12 at 19:30
@Hans Engler: Yes, this is exactly what I want, a normal distribution for which (n,f(n))= $(x_{n},y_{n})$. How is this accomplished? –  Ricky T Dec 23 '12 at 1:14
I asked an "either or" question and you responded "yes", so I am confused now. Which data are in your opinion approximately normally distributed? –  Hans Engler Dec 23 '12 at 1:46
As @Hans said. Also, your equation $(n,f(n))=(x_n,y_n)$ is confusing -- does that mean that $x_n=n$? In that case, why did you introduce the $x_n$ in the first place? Also, this equation appears to imply that you're considering a one-dimensional function. In that case, I wonder whether you're actually simply trying to fit a normalized Gaussian to a set of data points and the whole distribution aspect is just a distraction. In any case, what do you mean by "the mean of the data set"? –  joriki Dec 23 '12 at 10:52
@Hans Engler: Oops, sorry. I did mean a bivariate normal distribution. Unfortunately, I am not sure how to use this to produce the desired result. –  Ricky T Dec 23 '12 at 14:29

You need also $\sum x y$, otherwise you would exclude all the normal distributions where there is dependence between $X$ and $Y$.

The normal distribution that best fits the data is obtained by maximum likelihood estimation. It is the one that has the mean and covariance matrix equal to the empirical mean and empirical covariance matrix corresponding your sums (normalized by $n$).

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Thanks, I will try this approach. Actually, I am just concerned with the distributions where y depends on x, that is where the points have the form (n,f(n)) –  Ricky T Dec 23 '12 at 14:29
@RickyT Even if you are interested in the conditional distribution of $Y$ on $X$, not considering $\sum x y$ is akin to putting it equal to zero and having a regression coefficient of 0. –  Learner Dec 24 '12 at 3:11
@Learner: I edited the question, perhaps I should have phrased it better. Same question, what if we did consider $\sum xy$? –  Ricky T Dec 26 '12 at 22:56
@RickyT Do you know matrix algebra? –  Learner Dec 27 '12 at 0:02

You have the sufficient statistics for $\mu_X, \mu_Y, \sigma^2_X$ and $\sigma^2_Y$ so you can calculate their estimates directly using $$\bar{x} = \frac{1}{n}\sum_{i = 1}^n x_i, \,\,\, \bar{y} = \frac{1}{n}\sum_{i = 1}^n y_i$$ for the sample means and $$s^2_x= \frac{1}{n-1} \sum_{i=1}^n\left(x_i - \bar{x} \right)^ 2 = \frac{\sum_{i=1}^nx_i^2}{n-1} - \frac{n\bar{x}^2}{n-1} \\ s^2_y= \frac{1}{n-1} \sum_{i=1}^n\left(y_i - \bar{y} \right)^ 2 = \frac{\sum_{i=1}^ny_i^2}{n-1} - \frac{n\bar{y}^2}{n-1}$$ for the sample variances. As others have mentioned, without $\sum{xy}$ you will not be able to estimate the covariance between $X$ and $Y$, which the regression tag in your question suggests you want.

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Thanks a lot for this, it seems to be along the lines of what I am looking for. How would it change if we knew the values of $\sum xy$? That is, if we were given $\sum xy$, would we be able to produce the desired result? What exactly do the sample means represent in this situation? –  Ricky T Dec 26 '12 at 22:56
If you have $\sum xy$ you can estimate the covariance between $X$ and $Y$ like this: $\tfrac{1}{n - 1} \sum xy - \bar{y} \sum x - \bar{x} \sum y + \bar{x}\bar{y}$. The sample means are the center of your estimated two-dimensional normal distribution. –  Zoë Clark Dec 30 '12 at 19:38

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