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The joint probability function

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.

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the joint probability distribution for X, Y, ... is a probability distribution that gives the probability that each of X, Y, ... falls in any particular range or discrete set of values specified for that variable. In the case of only two random variables, this is called a bivariate distribution, but the concept generalizes to any number of random variables, giving a multivariate distribution.

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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

$X╲Y $ 0 1 2 3 Total
0 0 c 2c 3c 6c
1 2c 3c 4c 5c 14c
2 4c 5c 6c 7c 22c
Total 6c 9c 12c 15c 42c

Since $∑p_i=1$

∴ 42c=1

∴ $c=\frac{1}{42}$

With this value the probability distribution is

$X╲Y $ 0 1 2 3 Total
0 0 $\frac1{42}$ $\frac2{42}$ $\frac3{42}$ $\frac6{42}$
1 $\frac2{42}$ $\frac3{42}$ $\frac4{42}$ $\frac5{42}$ $\frac{14}{42}$
2 $\frac4{42}$ $\frac5{42}$ $\frac6{42}$ $\frac7{42}$ $\frac{22}{42}$
Total $\frac6{42}$ $\frac9{42}$ $\frac{12}{42}$ $\frac{15}{42}$ 1

∴ The marginal probability distributions of X & Y are:

X P(X)
0 $\frac6{42}$
1 $\frac{14}{42}$
2 $\frac{22}{42}$
X P(Y)
0 $\frac{6}{42}$
1 $\frac9{42}$
2 $\frac{12}{42}$
3 $\frac{15}{42}$

$E(X)=∑p_i x_i$

    =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

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

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

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

$ρ_{xy}=0.177$

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