Probability and Statistics independent random variables 63 L

Probability and Statistics, independent random variables.

6.3. Let the random variables X, Y, and Z be independent random variables. Find the following probabilities in terms of FX(x), FY(y), and FZ(z). (a) P(|X|

Solution

Note:- x^- means a number which is very close to x but still less than x for eg- 5^- means 4.9999999999999999999999.....

I have used it in my solution because P(x<=2) is F_x(2) but P(x<2) is F_x?(2^-) , hope u understand

P(x,y,z)=P(x)*P(y)*P(z) b/c they are independent

P(|x|<5)= P(-5<x<5)= F_x(5^-)-F_x(-5)

P(y<4)= F(4^-)

P(Z³>8)=P(Z>2) = 1-P(Z<=2) = 1-F(2)

P(|x|<5, y<4, Z³>8) = P(|x|<5)* P(y<4)*P(Z³>8) because they are independent

= (F_x(5^-)-F_x(-5))*F(4^-)*(1-F(2))

b) P(x=5, y<0, z>1) = P(x=5)*P(y<0)*P(z>1)

now, P(x=5)= F_x(5)-F_x(-5^-) , P(y<0) = F_y(0^-) , P(z>1)=1-P(z<=1) = 1- F_z(1)

so, P(x=5, y<0, z>1) = P(x=5)*P(y<0)*P(z>1)= (F_x(5)-F_x(-5^-))*( F_y(0^-) * (1- F_z(1))

c) P(min(x,y,z)<2) =1- P(x>=2,y>=2,z>=2) =1-P(x>=2)*P(y>=2)*P(z>=2) =1 -( (1- F_x(2^-))* (1-F_y(2^-?))* (1-F_z(2^-?)))

d) P(max(x,y,z)<6) =  P(x<6,y<6,z<6) = P(x<6)*P(y<6)*P(z<6) = F_x(6^-)* F_y(6^-?)* F_z(6^-?)