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I know that the variance of the difference of two correlated variables, $Y_1$ and $Y_2$ , is what the below formula shows, requiring $r$ the correlation between the two variables.

But suppose I have an amply large random sample of each of above variables, $Y_1^*$ and $Y_2^*$, in isolation.

Does the variance of $Y_2^* - Y_1^*$ give me a good estimate of the variance of $Y_2 -Y_2$?

enter image description here

Note: In this formula $V_1$ and $V_2$ are the variances of $Y_1$ and $Y_2$, respectively.

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  • $\begingroup$ Cannot understand what "an amply large random sample of each variables, Y1∗ and Y2∗." and "variance of Y2∗−Y1∗" mean. $\endgroup$
    – user158565
    Commented Aug 3, 2019 at 17:34
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    $\begingroup$ If you don't observe $Y_1$ and $Y_2$ together in pairs, you have no way of estimating their covariance or the variance of their difference. $\endgroup$
    – Jarle Tufto
    Commented Aug 3, 2019 at 21:16
  • $\begingroup$ @JarleTufto, thanks a lot! But if I do observe $Y_1$ and $Y_2$ in pairs, then I can simply subtract one from the other, and compute the variance of their difference? $\endgroup$
    – rnorouzian
    Commented Aug 3, 2019 at 21:41
  • $\begingroup$ What you propose is reasonable. For one possible implementation in R, please see my Answer. [Lacking definitions of $V_1, V_2,$ I can make no sense of the equation in the 'picture' at the end of your Question. But we don't need it.] $\endgroup$
    – BruceET
    Commented Aug 3, 2019 at 21:46
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    $\begingroup$ @BruceET, thank you, $V_1$ and $V_2$ are the variances of $Y_1$ and $Y_2$, respectively. $\endgroup$
    – rnorouzian
    Commented Aug 3, 2019 at 22:20

1 Answer 1

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Theoretical results. First, an example with results from some theoretical formulas. Suppose $X_1 \sim \mathsf{Norm}(\mu = 50, \sigma=7),\,$ $X_2 \sim \mathsf{Norm}(40, 5),$ and $W \sim \mathsf{Norm}(0, 3).$

Then let $Y_1 = X_1 + W,\, Y_2 = X_2 + W$ so that $$Cov(Y_1,Y_2) = Cov(X_1+W, X_2+W)\\ = Cov(X_1,X_2)+Cov(X_1,W) +Cov(W,X_2)+Cov(W,W)\\ = 0+0+0+Cov(W,W) = Var(W) = 9$$ because $X_1, X_2,$ and and $W$ are mutually independent. Moreover, by independence, $Var(Y_1) = Var(X_1) + Var(W) = 7^2 + 3^2 = 58$ and, similarly, $V(Y_2) = 34,$ so that $$Var(Y_1 - Y_2) = Var(Y_1) + Var(Y_2) - 2Cov(Y_1, Y_2) = 58 + 34 - 2(9) = 74.$$

Approximation by simulation. If we simulate a million realizations each of $X_1, X_2,$ and $W$ in R, then we can approximate some key quantities from the theoretical results. [R parameterizes the normal distribution in terms of $\mu$ and $\sigma.]$

With a million iterations, it is reasonable to expect approximations accurate to three significant digits for standard deviations and about two for variances and covariances. [The weak law of large numbers promises convergence, the central limit theorem allows computations of margin of simulation error based on sample size.]

set.seed(2019);  m = 10^6
w = rnorm(m, 0, 3)
x1 = rnorm(m, 50, 7);  x2 = rnorm(m, 40, 5)
sd(w); sd(x1); sd(x2)
[1] 3.003595    # aprx SD(W) = 3
[1] 6.994074    # aprx SD(X1) = 7
[1] 4.997491    # aprx SD(X2) = 5

y1 = x1 + w;  y2 = x2 + w
cov(y1,y2);  cor(y1,y2)
[1] 9.028998    # aprx Cov(Y1,Y2) = 9
[1] 0.2033905
sd(y1);  sd(y2); sd(y1 - y2)
[1] 7.61474
[1] 5.829802
[1] 8.597258
var(y1); var(y2);  var(y1 - y2)
[1] 57.98426    # aprx V(Y1) = 58
[1] 33.98659    # aprx V(Y2) = 34
[1] 73.91285    # aprx V(Y1 - Y2) = 74

Note: In case you don't know the 'back story' involving random variables $X_1, X_2,$ and $W,$ but you do know the correlation $\rho$ between $Y_1$ and $Y_2,$ here is a way to simulate two normal random variables with a specified correlation $\rho.$ (See this Q&A for discussion.)

If $Z \sim \mathsf{Norm}(0,1)$ and, independently $Y_2 \sim \mathsf{Norm}(0,1),$ then random variables $Y_1 = \rho Y_2 + \sqrt{1-\rho^2}Z$ and $Y_2$ have correlation $\rho.$

set.seed(804);  n = 10^6;  rho = .8
z = rnorm(n);  y2 = rnorm(n)
y1 = rho*y2 + sqrt(1-rho^2)*z
mean(y1);  sd(y1)
[1] 0.0005254835  # aprx E(Y1) = 0
[1] 0.9987712     # aprx SD(Y1) = 1
cor(y1,y2)
[1] 0.7998153     # aprx rho = Cor(Y1. Y2) = 0.8
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  • $\begingroup$ Yes. It would be best to have a large sample from the populations of interest. Failing that, you can try to guess parameters and get corresponding answers from a large simulation. $\endgroup$
    – BruceET
    Commented Aug 4, 2019 at 0:47
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    $\begingroup$ By "guess parameters", are you only referring to $r$, or also $V_1$ and $V_2$, the variances of the actual random variables? Also, what do you think about the comment by @ Jarle Tufto who stated: If you don't observe $Y_1$ and $Y_2$ together in pairs, you have no way of estimating their covariance or the variance of their difference? $\endgroup$
    – rnorouzian
    Commented Aug 4, 2019 at 0:57
  • $\begingroup$ Yes, in particular the population standard deviations 3, 7, and 5 in the 2nd and 3rd lines of code. The correlations are inherited from them. $\endgroup$
    – BruceET
    Commented Aug 4, 2019 at 1:01
  • $\begingroup$ Yes, but think before you simulate. For the distribution of the mean of independent observations, standard theory may tell you all you need to know. $\endgroup$
    – BruceET
    Commented Aug 4, 2019 at 1:13
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    $\begingroup$ Dear Bruce, I thought you might be interested in THIS QUESTION. $\endgroup$
    – rnorouzian
    Commented Aug 5, 2019 at 1:30

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