Does independence between random variables imply independence between related events?
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Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?
If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?
random-variable independence
asked Nov 22 at 11:39
mickkk
374314
add a comment |
up vote
1
down vote
favorite
Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?
If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?
random-variable independence
asked Nov 22 at 11:39
mickkk
374314
See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?
If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?
random-variable independence
asked Nov 22 at 11:39
mickkk
374314
Say I have two random variables X1 and X2 and that they are independent. Am I guaranteed that the events "X1 is less than x1" and "X2 is less than x2" are independent?
If not, under which conditions is this the case? Or better, what is a sufficient condition for having independence between those two events?
random-variable independence
random-variable independence
asked Nov 22 at 11:39
mickkk
374314
asked Nov 22 at 11:39
mickkk
374314
edited Nov 22 at 11:53
asked Nov 22 at 11:39
mickkk
374314
asked Nov 22 at 11:39
mickkk
374314
asked Nov 22 at 11:39
mickkk
374314
374314
See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55
add a comment |
See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55
See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55
See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55
add a comment |
2 Answers
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I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.
Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.
answered Nov 22 at 15:01
user158565
5,1451318
add a comment |
up vote
1
down vote
If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
$$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
But, $$F_{X,Y}(a,b)
stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
while $$F_{X}(a)
stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
and so $(*)$ is saying that
$$P(Acap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.
Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.
answered Nov 22 at 15:01
user158565
5,1451318
add a comment |
up vote
2
down vote
I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.
Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.
answered Nov 22 at 15:01
user158565
5,1451318
add a comment |
up vote
2
down vote
up vote
2
down vote
I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.
Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.
answered Nov 22 at 15:01
user158565
5,1451318
I remember that the definition of two random variables $X_1$ and $X_2$ are independent is for any event generated by random variables $X_1$ and event generated by $X_2$ are independent. So events $(X_1<x_1)$ and $(X_2<x_2)$ are independent is the condition that two random variables $X_1$ and $X_2$ are independent.
Maybe you need to check the textbook or search the internet to find the definition of independent of two random variables.
answered Nov 22 at 15:01
user158565
5,1451318
answered Nov 22 at 15:01
user158565
5,1451318
answered Nov 22 at 15:01
user158565
5,1451318
answered Nov 22 at 15:01
user158565
5,1451318
5,1451318
add a comment |
add a comment |
up vote
1
down vote
If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
$$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
But, $$F_{X,Y}(a,b)
stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
while $$F_{X}(a)
stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
and so $(*)$ is saying that
$$P(Acap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
add a comment |
up vote
1
down vote
If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
$$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
But, $$F_{X,Y}(a,b)
stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
while $$F_{X}(a)
stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
and so $(*)$ is saying that
$$P(Acap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
add a comment |
up vote
1
down vote
up vote
1
down vote
If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
$$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
But, $$F_{X,Y}(a,b)
stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
while $$F_{X}(a)
stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
and so $(*)$ is saying that
$$P(Acap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
If $X$ and $Y$ are independent random variables, then it is always the case that the events $A = {X leq a}$ and $B = {Y leq b}$ are independent events. Specifically, one of the (equivalent) definitions of independence of two random variables is that the joint CDF factors into the product of the individual (a.k.a. marginal) CDFs. That is, we are insisting that independence of $X$ and $Y$ means that
$$F_{X,Y}(a,b) = F_X(a)F_Y(b)~text{for all real numbers}~a~text{and}~btag{*}$$
But, $$F_{X,Y}(a,b)
stackrel{Delta}{=} Pleft({X leq a, Y leq b}right) = Pleft({Xleq a}cap {Y leq b}right) = P(Acap B)$$
while $$F_{X}(a)
stackrel{Delta}{=} Pleft({X leq a}right) = P(A), quad F_{X}(b)
stackrel{Delta}{=} Pleft({Y leq b}right) = P(B)$$
and so $(*)$ is saying that
$$P(Acap B) = P(A)P(B),$$
that is, $A$ and $B$ are independent events.
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
answered Nov 22 at 15:03
Dilip Sarwate
29.6k252146
29.6k252146
add a comment |
add a comment |
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See stats.stackexchange.com/questions/94872/… inter alia.
– whuber♦
Nov 22 at 14:55