1
$\begingroup$

Given is a two-dimensional random variable $(X,Y)$. Here is the table of single probabilities $p_{ij}$ of $(X,Y)$ which was incomplete and I filled it with needed values so it's correct and complete.

\begin{array}{r|rrrr|r} X\setminus Y & 1 & 2 & 3 & 4 & p_i \\ \hline -1 & 0 & 0.01 & 0.09 & 0.10 & 0.2 \\ 0 & 0.6 & 0.03 & 0 & 0.07 & 0.7 \\ 1 & 0 & 0.06 & 0.01 & 0.03 & 0.1 \\ \hline p_j & 0.6 & 0.1 & 0.1 & 0.2 & 1 \\ \end{array}

Now I want calculate correlation between $X$ and $Y$ but not know how do it correctly? Because it can be asked in exam and I don't see example for it.

When I understand it correct I need find expected values of $X$ and $Y$, e.g. $E(X)$ and $E(Y)$ then

$$\mathrm{correlation}(X,Y)=E(XY) - E(X) \cdot E(Y).$$

Assuming this is correct, I still have no idea how compute these expected value from the table :(

$\endgroup$
  • $\begingroup$ Do you know how to compute $E(X)$ from such a table? If so, why don't you add that to your question, clicking on "edit" beneath the question to show us how you'd do that. $\endgroup$ – John Hughes Jan 13 '18 at 14:01
  • 1
    $\begingroup$ Firstly note that $\mathbb E(XY)- \mathbb E(X)\cdot \mathbb E(X)$ is the $\texttt{covariance}$, not the $\texttt{correlation}$ coefficient. The correlation coefficient is $\rho_{XY}=\frac{Cov(X,Y)}{\sigma_X\cdot \sigma_Y}$ $\endgroup$ – callculus Jan 13 '18 at 14:09
  • 1
    $\begingroup$ I have tried to improve the readability of your question by introducing $\rm \LaTeX$. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 13 '18 at 14:09
  • $\begingroup$ @GNUSupporter Thank you it look great I didn't know how do it in latex!!! $\endgroup$ – roblind Jan 13 '18 at 15:12
3
$\begingroup$

We know that $$E(X) = \sum xp(x) = - 0.2+0+0.1=-0.1$$ $$E(X^2) = 0.2+0+0.1=0.3$$ $$E(Y)= \sum yp(y) = 0.6+0.2+0.3+0.8=1.9$$ $$E(Y^2) = 0.6+0.4+0.9+3.2=5.1$$ $$E(XY) = \sum xyp(x, y) = - 0.18$$ Hence, $$\operatorname{Cov} (X, Y) = E(XY) - E(X) E(Y) = 0.01$$ $$\sigma_x^2 = E(X^2)-[E(X)]^2 = 0.29$$ $$ \sigma_y^2 = E(Y^2)-[E(Y)]^2 = 1.49$$ Therefore, $$r_{XY} = \frac{0.01}{\sqrt{1.49\times 0.29}} = 0.015$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.