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(X2), E(Y2), var(X), var(Y), cov(X, Y) and ρ.

**Mumbai University > Electronics and Telecommunication Engineering > Sem 5 > Random Signal Analysis

Marks: 10M

Year: May 2016

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=1/42

With this value the probability distribution is

∴The marginal probability distributions of X & Y are:

E(X)= $∑{p_i} {x_i}$

=0+1×14/42+2×22/42

E(X)=58/42=1.381

E(Y)=$∑{p_i}{ y_i}$

=0+1×9/42+2×12/42+3×15/42

E(Y)=78/42=1.857

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

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

=0+1(0×2/42+1×3/42+2×4/42+3×5/42)+2(0×4/42+1×5/42+2×6/42+3×7/42)

=26/42+76/42

E(XY)=102/42=2.429

E($X^2$ )=$∑{p_i} {x_i^2}$

=0+1×14/42+2^2×22/42

E($X^2$ )=102/42=2.429

E($Y^2$ )=$∑{p_i} {y_i^2}$

=0+1×9/42+$2^2$×12/42+$3^2$×15/42

E($Y^2$ )=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

$σ_x$=√(Var(X) )=√0.5218=0.7224

$σ_y$=√(Var(Y) )=√1.1226=1.06

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