written 7.9 years ago by
teamques10
★ 64k
|
• modified 4.3 years ago |
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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written 7.9 years ago by
teamques10
★ 64k
|
• modified 7.9 years ago |
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
$σ_x=\sqrt{(Var(X))}=\sqrt{0.5218}=0.7224$
$σ_y=\sqrt{(Var(Y) )}=\sqrt{1.1226}=1.06$
$ρ_{xy}=\frac{Cov(X,Y)}{(σ_x σ_y )}=-\frac{0.1355}{(0.7224*1.06)}=0.177$
$ρ_{xy}=0.177$
and 2 others joined a min ago.
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.