The joint probability function of two discrete r.v's X and Y is given by f(x, y) = c(2x+y), where x and y can assume all integers such that 0 ≤ x ≤2,0 ≤ y ≤ 3 and f(x,y) =0 otherwise. Find E(X), E(Y) , E(XY), E($X^2$), E($Y^2$), var(X), var(Y), cov(X, Y) and ϱ.

X and Y are discrete RVs

To find c: We can tabulate the probabilities as follows:

$f(x,y)= c(2x+y)$         $0 ≤ x ≤ 2,0 ≤ y ≤ 3$

        =0

Since $∑p_i=1$

∴ 42c=1

∴ $c=\frac{1}{42}$

With this value the probability distribution is

∴ The marginal probability distributions of X & Y are:

$E(X)=∑p_i x_i$

    =0+1×$\frac{14}{42}+2×\frac{22}{42}$

$E(X)=\frac{58}{42}=1.381$

$E(Y)=∑p_i y_i$

    $=0+1×\frac9{42}+2×\frac{12}{42}+3×\frac{15}{42}$

$E(Y)=\frac{78}{42}=1.857$

$E(XY)=∑_{i=0}^2∑_{j=0}^3$ $x_i y_j$ p(x=i,y=j)

$=0*0+1.∑_{j=0}^3 y_j p(x=1,y=j)+2.∑_{j=0}^3y_j p(x=2,y=j)$

$=0+1*(0×\frac2{42}+1×\frac3{42}+2×\frac4{42}+3×\frac5{42}+2*(0×\frac4{42}+1×\frac5{42}+2×\frac6{42}+3×\frac7{42}$

$=\frac{26}{42}+\frac{76}{42}$

$E(XY)=\frac{102}{42}=2.429$

$E(X^2 )=∑p_i x_i^2$

    $=0+1×\frac{14}{42}+2^2×\frac{22}{42}$

$E(X^2 )=\frac{102}{42}=2.429$

$E(Y^2 )=∑p_i y_i^2$

=$0+1×\frac9{42}+2^2×\frac{12}{42}+3^2×\frac{15}{42}$

$E(Y^2 )=\frac{192}{42}=4.571$

Var$(X)=E(X^2 )-{E(X)}^2$

    $=2.429-1.381^2$

   $=2.429-1.9072$

Var(X)=0.5218

$Var(Y)=E(Y^2 )-{E(Y)}^2$

    $=4.571-1.857^2$

    =4.571-3.4484

Var(Y)=1.1226

cov(X,Y)=E(XY)-E(X)E(Y)

    =2.429-1.381*1.857=2.429-2.565=-0.1355

Cov(XY)=-0.1355

$σ_x=\sqrt{(Var(X))}=\sqrt{0.5218}=0.7224$

$σ_y=\sqrt{(Var(Y) )}=\sqrt{1.1226}=1.06$

$ρ_{xy}=\frac{Cov(X,Y)}{(σ_x σ_y )}=-\frac{0.1355}{(0.7224*1.06)}=0.177$

$ρ_{xy}=0.177$