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The question states:

Consider a set of random variables $X_i$, where $i=1,...n$. Each $X_i$ is normally distributed with mean $0$ and variance $1$, i.e. $X_i$ are $\mathcal N(0,1)$. What is the mean and the variance of the random variable $Y$, where $Y=X_1+...+X_n$.

How do I do this?

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    $\begingroup$ google.be/… $\endgroup$ – ocram May 4 '13 at 20:46
  • $\begingroup$ So Y is my sum of normally distributed RVs and I should find the mean and variance of that, correct? $\endgroup$ – Skytbest May 4 '13 at 20:55
  • $\begingroup$ If you can't think of an analytic solution, you could try simulating it. That may spur you to find the analytic solution. $\endgroup$ – Peter Flom - Reinstate Monica May 4 '13 at 21:13
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Regarded as a question in probability theory, the answer to this question is that $$E[Y]=E[X_1+X_2+\cdots +X_n] = E[X_1] + E[X_2]+\cdots + E[X_n]$$ via a result known as the linearity of expectation, and since the random variables all have zero mean in this particular instance, $E[Y]=0$. On the other hand,

the information given is insufficient to determine the variance of $Y$.

The variance of $Y$ is $$\operatorname{var}(Y) = \sum_{i=1}^n \operatorname{var}(X_i) + 2 \sum_{i=1}^{n-1}\sum_{j=i+1}^n\operatorname{cov}(X_i,X_j)$$ and so unless one knows (or makes assumptions about) the covariances, the variance of $Y$ cannot be determined. One common assumption is that the $X_i$ are independent random variables in which case the variance of $Y$ is just the sum of the variances. The weaker condition that the $X_i$ are uncorrelated random variables also leads to the same result, which in this instance is that $\operatorname{var}(Y)=n$.

The answers are different if one is talking about the sample mean and the sample variance of $n$ samples from a standard normal distribution.

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  • $\begingroup$ Are you sure the variance cannot be found? I was just looking on the wiki page for 'Sum of normally distributed random variables' and it said "This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances". So can we not just add the variances together in my example? $\endgroup$ – Skytbest May 6 '13 at 15:14
  • $\begingroup$ Yes, I am sure that the variance cannot be found from the information in the current version of your question. Note that the Wikipedia page that you refer to specifically says that the two random variables are independent which word is nowhere to be found in your question. In my answer, I did point out that if one assumes that the random variables are independent, then the variances can be added up. Your $Y \sim N(0,n)$, but normality is irrelevant to the issue of the variances adding. Variances add if the variables are uncorrelated, including as a special case, independent. $\endgroup$ – Dilip Sarwate May 6 '13 at 15:23
  • $\begingroup$ Ah, I see your point. I think I'll write above that I'm assuming the RVs are independent otherwise this cannot be done. I don't think the professor would be trying to throw us for a loop like this. Thank you! $\endgroup$ – Skytbest May 6 '13 at 15:37
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Working off Peter Flom's suggestion, look at this R code:

x <- as.data.frame(matrix(rnorm(50000,0,1), nr = 5000, nc = 10))
x$y <- x$V1 + x$V2 + x$V3 + x$V4 + x$V5 + x$V6 + x$V7 + x$V8 + x$V9 + x$V10

We generate 10 variables (V1 through V10), each with a mean of 0 and a standard deviation of 1. Y is constructed to be the sum of the ten variables for each observation. You can then run:

colMeans(x)
diag(var(x))

And those results should give you some insight about what to expect.

Then, look at this Wikipedia entry (the formula for Z specifically) and see if your insights match theirs!

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