# Let the random variables X and Y have the joint pmf f(x,y)=(x+y)/32, x=1,2, and y=1,2,3,4. What are the means mu_x and mu_y, the variances sigma ""_x^2 and sigma "" _y^2, and the correlation coefficient rho?

Apr 25, 2016

${\mu}_{x} = 1.5625$
${\mu}_{y} = 2.8125$
${\sigma}_{x}^{2} = 0.24609375$
${\sigma}_{y}^{2} = 1.15234375$
$\rho \approx - 0.0366766$

#### Explanation:

Let's calculate those joint probabilities:

e.g. $P \left(X = 2 \wedge Y = 3\right) = 0.15625$.
It is now quite easy to obtain separate probability distributions of $X$ and $Y$: we should sum rows or columns of area B2:E3, respectively, e.g.
$P \left(X = 2\right) = {\sum}_{i = 1}^{4} P \left(X = 2 \wedge Y = i\right) = 0.09375 + \ldots + 0.1875 = 0.5625$
Now we can use known formulas for the values we're looking for:
${\mu}_{x} = \text{E} X = 1 \cdot 0.4375 + 2 \cdot 0.5625 = 1.5625$
${\mu}_{y} = \text{E} Y = 1 \cdot 0.15625 + 2 \cdot 0.21875 + 3 \cdot 0.28125 + 4 \cdot 0.34375 = 2.8125$
$\text{E} {X}^{2} = {1}^{2} \cdot 0.4375 + {2}^{2} \cdot 0.5625 = 2.6875$
$\text{E} {Y}^{2} = {1}^{2} \cdot 0.15625 + \ldots + {4}^{2} \cdot 0.34375 = 9.0625$
${\sigma}_{x}^{2} = {\text{E"X^2-"E}}^{2} X = 2.6875 - {1.5625}^{2} = 0.24609375$
${\sigma}_{y}^{2} = {\text{E"Y^2-"E}}^{2} Y = 9.0625 - {2.8125}^{2} = 1.15234375$

Now for the correlation coefficient $\rho$ we need the covariance ${\sigma}_{x y}$ so let's find the probability distribution of $X Y$: (here ${p}_{i j} = P \left(X = i \wedge Y = j\right)$ )

$P \left(X Y = 1\right) = {p}_{11} = 0.0625$
$P \left(X Y = 2\right) = {p}_{12} + {p}_{21} = 0.1875$
$P \left(X Y = 3\right) = {p}_{13} = 0.125$
$P \left(X Y = 4\right) = {p}_{14} + {p}_{22} = 0.28125$
$P \left(X Y = 5\right) = 0$
$P \left(X Y = 6\right) = {p}_{23} = 0.15625$
$P \left(X Y = 7\right) = 0$
$P \left(X Y = 8\right) = {p}_{24} = 0.1875$

Now, for the covariance:
$\text{E} \left(X Y\right) = 1 \cdot 0.0625 + \ldots + 8 \cdot 0.1875 = 4.375$
${\sigma}_{x y} = \text{E"(XY)-"E"X*"E} Y = - 0.01953125$
$\rho = \frac{{\sigma}_{x y}}{\sqrt{{\sigma}_{x}^{2} \cdot {\sigma}_{y}^{2}}} = - 0.0366765706779718$