On pg. 44 of Introduction to Statistical Learning:

Here we create two correlated sets of numbers, x and y, and use the cor() function to compute the correlation between them.

x = rnorm(50)
y = x + rnorm(50, mean=50, sd=.1)
[1] 0.995

Observation: Playing around with this and related examples, I see the following:

Let $x$ be vector of length $n$ randomly sampled from a normal population with mean $0$ and standard deviation $1$. Let $y = x + z$ such that $z$ is a vector of length $n$ with sample mean $\bar{z}$ and sample standard deviation $s_z$.

  1. Then $x$ and $y$ tend to be correlated so long as $s_z$ is reasonably low (the lower it is, the more $x$ and $y$ tend to be correlated).

  2. I have also noticed that the sample mean of $z$ doesn't tend to affect the correlation between $x$ and $y$.

Question: Formally, why do (1) and (2) hold? (Really, I'm assuming that an understanding of (1) will automatically clarify (2)).

  • 1
    $\begingroup$ See our threads on computing covariance of sums and apply the definition of correlation. That gives a rigorous algebraic approach. Alternatively, drawing a picture of how $y$ is constructed from $x$ and $z$ will make the result immediately obvious. $\endgroup$
    – whuber
    Commented May 27, 2016 at 20:51

1 Answer 1


For illustration purpose, consider smaller vectors:

x = rnorm(5) # x = -0.3260365  0.5524619 -0.6749438  0.2143595  0.3107692

First, let's say $z$ is a vector of constant values, i.e., $s_z = 0$

z <- rep(50, 5)
y <- x + z

Now plot(x,y) will produce a perfect straight line and cor(x,y) will return 1 meaning perfect correlation between $x$ and $y$. It is because given an $x$ value, we can accurately predict $y$ value ($y = x + 50$).
Again, take random values of $z$ with mean $\bar z = 50$ and standard deviation $s_z = 5$. What this means is that now $y$ depends not only on the randomness of $x$, but also randomness of $z$. Larger SD of $z$ indicates the values are more distant from the mean, $50$. So if we try to predict $y$ from $y = x + 50$, as we did before, our prediction will be a lot less accurate meaning weaker correlation. Speaking of your second question, as it should have been clear by now, SD instead of mean affects the correlation.


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