no

Statistics
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Question 1
Bivariate discrete random distribution is a function that examines the behavior of two random variables. (Stepanov, 2014). They can be either dependent variables or independent random variables. Let P (x) and F (y) be marginal distribution of x and y respectively and let f (x, y) be their joint distribution function,
The e and y are said to be independent random variables if and on if p (x)*f (y)=f (x, y) CITATION NBa14 l 1033 (Stepanov, 2014)a) Compute the following probabilities.
I) Prob (y<2)
Prob (y<2)= probe(y=0) and probe(y=1)
= probe(y=0,) +probe(y=1)
ii) probe[y<2,x>0]
probe[y<2,x>0]= probe[y=0,x=1]+ probe[y=1,x=0]+ probe[y=0,x=2]+ probe[y=1,x=2]
=0.21+0.11+0.08+0.15
=0.55
b) the marginal probability density functions for x and y.
let f(x,y) be the joint probability of x and y and let g(x) and h(x) be the marginal probabilities of x and y respectively.
g(x)=  ∑all y (x,Y) h(y)=  ∑all x f(x,y)
X 0 1 2
G(x) 0.18 0.51 0.31
Y 0 1 2
H(y) 0.34 0.36 0.3
C) The mean and variance of X.
Let x̅ be the mean σ2 be the variance of x
x̅=E(x)
=∑x.p(x)
=(0*0.18)+(1*0.51)+(2*0.31)
=1.13
 σ2=E(x2)+E(x)-[E(x)]2
={ (02*0.18)+(12*0.51)+(22*0.31)}+ 1.13-(1.13)2
=1.75+1.13-1.2769
=1.6031
d) conditional probability of y given x+1 and the conditional mean and variance of y given x=1
let P(y) be the conditional probability of y given x=1
p(y)=0.1+0.11+01.5
=0.36
let x̅ be the conditional mean of y given x=1
x̅=yP(y) when x=1
=(0*0.

1)+(1*0.11)+(2*0.15)
=0.41
The variance of y given x=1
Var(x)= E(y2)+E(y)-[E(y)]2 ,when x=1
={ (02*0.1)+(12*0.11)+(22*0.15)}+ 0.36-(0.36)2
=0.8404
e) covariance of x and y
cov(x,y)=(XY)-E(x)E(y)
={(0*0.05)+(0*0.1)+(0*0.03)+(0*0.21)+(1*0.11)+(2*0.19)+(0*0.08)+(2*0.15)+(4*0.08)}-{1.13*0.96}
=1.51-1.0848
=0.4252
f) X and y are not independent this is because
g(x).h(y) ≠f(x,y)
for example
g(x=1).h(y=1) =0.51*0.36
=0.1836 ≠ 0.11
References
BIBLIOGRAPHY Stepanov, N. B. (2014, March). On the Use of Bivariate Mellin Transform in Bivariate Random Scaling and Some Applications.