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I am having coding the following in R I want to pick 10 random numbers from a standard normal distribution whose sum equals 5. I have the following code so far (below), but this returns "numeric(0)" when the random numbers don't satisfy the condition. What I want is to choose 10 random numbers that do satisfy this condition. Is there a way to re-scale these numbers once they have been picked, or can I somehow insert a condition for this into "rnorm" ? Help very much appreciated!

a <- rnorm(10, 0, 1)

ones <- matrix(1, nrow=10, ncol=1 )

A <- a[t(a) %*% ones == 5]

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The way I interpret this question, we are asking to generate from the distribution $(X_1, \ldots, X_{10} \mid \sum_{i = 1} ^ {10} X_i = 5)$ where $X_1, \ldots, X_n \stackrel {iid} \sim \mathcal N(0,1)$. This is easy enough - the joint distribution of $(X_1, \ldots, X_9, \sum_i X_i)$ is multivariate normal with mean vector $\mu = (0,\ldots,0)$ and covariance matrix

$$\Sigma = \pmatrix{1 & 0 & \ldots & 0 & 1 \\ 0 & 1 & \ldots & 0 & 1 \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & 1 \\ 1 & 1 & \ldots & 1 & 10}.$$

From this, it can be shown that $(X_1, \ldots, X_9 \mid \sum_i X_i)$ is, again, a normal distribution with mean $\mu^\star = (\bar X, \bar X, \ldots, \bar X)$ and covariance matrix $\Sigma^\star = \mathbf I - \frac 1 {10} \mathbf J$ where $\mathbf J$ is a matrix with all entries equal to $1$ - see, for example, here. So, to do this in R,

library(MASS)
Sigma  <- diag(9) - .1
X      <- numeric(10)
X[1:9] <- mvrnorm(1, rep(5 / 10, 9), Sigma)
X[10]  <- 5 - sum(X[1:9])

And now $(X_1, \ldots, X_{10})$ is drawn from the distribution $(X_1, \ldots, X_{10} \mid \sum_i X_i = 5)$.

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    $\begingroup$ +1 for interpreting the question in a constructive manner and providing a good approach. $\endgroup$
    – whuber
    May 12 '14 at 14:23
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You can't actually do this. By adding the constraint of sum=5, you're reducing the degrees of freedom by 1 and therefore eliminating true randomness. You can either 1) pick 9 truly random numbers from a distribution, followed by one fixed number to get to a sum of 5, or 2) approximate sum=5 by doing a random sample from a distribution with a fixed mean and small variance, as Peter Flom has already explained.

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You can generate 10 identically distributed normal random variables that sum to 5. But they won't be standard normal and they won't be independent.

Generate $X_1 \sim N({1 \over 2},1).$ Let $X_2 = 1-X_1.$ Generate $X_3 \sim N({1 \over 2},1).$ Let $X_4 = 1-X_3.$ And so on, generating 5 pairs of dependent random variables, whose sum is 5.

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10 random numbers from a standard normal will have mean 0 and thus the sum will be about 0, not 5. If you ignore the "standard" part, you will still not be able to pick 10 random numbers from a normal distribution whose sum is exactly 5. You can get 10 such numbers with sum very close to 5 by e.g.

set.seed(123)
vars <- rnorm(10, 0.5, 0.001)
sum(vars)
vars

and if you make the sd (0.001) even smaller then the sum will tend to be even closer to 5.

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  • $\begingroup$ Oops, my bad. I will correct $\endgroup$
    – Peter Flom
    May 11 '14 at 11:09

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