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$X,Y,Z$ are IID ${\rm Poisson}(\lambda)$. Find the correlation between $X$ and $X+Y+Z$.

So far I've tried making a scatterplot, but it did not help.

I also tried looking at this figure from Wikipedia

enter image description here
(image released to public domain)

and got even more confused. Should I be using this formula from Wikipedia

\begin{align} r_{xy} &=\frac{\sum x_iy_i-n \bar{x} \bar{y}}{n s_x s_y} \\ &=\frac{n\sum x_iy_i-\sum x_i\sum y_i}{\sqrt{n\sum x_i^2-(\sum x_i)^2}~\sqrt{n\sum y_i^2-(\sum y_i)^2}}?????? \end{align}

If so, what is $x_i, y_i$ and $\bar{x}, \bar{y}$? And $n$?

What about the rank correlation coefficient? Is that applicable here??

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    $\begingroup$ SAMPLE correlation is very different from POPULATION correlation. You are confusing the two, and the answer below does not mention this $\endgroup$ – Taylor Feb 1 '17 at 16:54
  • $\begingroup$ @Taylor, can you explain what you mean? Can't you look at a scatter plot to quantify correlation? That's what I was taught in my intro stats class, or to use SPSS, but this upper level stats class has proven much harder! :-( $\endgroup$ – casecontrol_logistic Feb 1 '17 at 19:40
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    $\begingroup$ sample correlation is something you calculate from your dataset. your data is random, so your sample correlation is random. and most of the time you don't know the exact probability distribution with real data. population correlation is something you calculate with your real probability distribution. It is a non-random number that doesn't rely on observations of your data...but it does rely on the probability distribution $\endgroup$ – Taylor Feb 1 '17 at 20:22
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If $X,Y$ and $Z$ are independent, then their covariance is 0. So is their correlation.

For the sums look at $$ cov(X,X+Y+Z) = Cov(X,X) + Cov(X,Y) + Cov(X,Z) $$ the latter 2 are zero (due to independence) and the first one is $Cov(X,X) = Var(X) = \lambda$.

For the variance we get $$ VAR(X+Y+Z) = VAR(X) + VAR(Y) + VAR(Z) = 3 \lambda $$ due to independence. Thus the correlations is $$ cor(X,X+Y+Z) = \frac{cov(X,X+Y+Z)}{\sqrt{VAR(X+Y+Z)} \sqrt{VAR(X)}} = \frac{\lambda}{\lambda \sqrt{3}} = 1/\sqrt{3}. $$

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  • $\begingroup$ Thank you but how did you get the first line? I didn't think covariances added like that? Also, how does this relate to the scatterplot or SPSS output you normally look at when calculating correlations? Why doesn't the correlation depend on $\lambda$ and what does the $1/\sqrt{3}$ in real world terms here? Still very confused :-( $\endgroup$ – casecontrol_logistic Feb 1 '17 at 19:45
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    $\begingroup$ Ultimately it follows from basic properties of expectation. Cov(X,X+Y) = E(X(X+Y)) - E(X)(E(X+Y). and we have linearity of expectation, so we get E(X*X)+E(XY)-E(X)E(X)-E(X)E(Y) = Cov(X,X)+ Cov(X,Y). $\endgroup$ – Glen_b Feb 2 '17 at 3:07
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    $\begingroup$ Right, as already pointed out, we speak about random variables here. This is a pure model and we can derive the result from the definitions. The sample estimators are one way to estimate these objects from real data. The plots are examples of special data and have noting to do with your case... $\endgroup$ – Richard Feb 2 '17 at 6:30

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