Question-86
Let \(X,Y\) \(\sim \text{Uniform}\{1,2\}\) be i.i.d discrete random variables. Define \(U=X+Y\) and \(V=X\). Find the correlation coefficient between \(U\) and \(V\). Enter your answer correct to two decimal places.
\(0.71\)
We have the covariance between \(U\) and \(V\) given as:
\[ E[ UV] -E[ U] E[ V] \]
So first we compute:
\[ \begin{aligned} E[ U] & =E[ X] +E[ Y] =2\times \frac{3}{2} =3\\ E[ V] & =E[ X] =\frac{3}{2} \end{aligned} \]
where we have used the linearity of expectation in the first equality. Next:
\[ \begin{aligned} E[ UV] & =E[( X+Y) X]\\ & =E\left[ X^{2} +XY\right]\\ & =E\left[ X^{2}\right] +E[ XY] \end{aligned} \]
Using the independence of \(X\) and \(Y\):
\[ \begin{aligned} E[ UV] & =E\left[ X^{2}\right] +E[ X] E[ Y]\\ & =\frac{5}{2} +\frac{9}{4}\\ & =\frac{19}{4} \end{aligned} \]
The covariance is therefore:
\[ \text{cov} =\frac{19}{4} -\frac{9}{2} =\frac{1}{4} \]
Next we note that \(\sigma _{X}^{2} =E\left[ X^{2}\right] -E[ X]^{2} =\frac{5}{2} -\frac{9}{4} =\frac{1}{4}\). Now for \(\sigma _{U}\) and \(\sigma _{V}\). For \(\sigma _{U}^{2}\), we use independence of \(X\) and \(Y\) again.
\[ \begin{aligned} \sigma _{U}^{2} & =\sigma _{X}^{2} +\sigma _{Y}^{2} =\frac{1}{2}\\ \sigma _{V}^{2} & =\frac{1}{4} \end{aligned} \]
Finally, the correlation coefficient is:
\[ \frac{\frac{1}{4}}{\frac{1}{2} \times \frac{1}{\sqrt{2}}} =\frac{\sqrt{2}}{2} \approx 0.71 \]