Does independence between random variables imply independence between related events?
$begingroup$
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 '18 at 11:39
mickkkmickkk
379314
$endgroup$
add a comment |
$begingroup$
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 '18 at 11:39
mickkkmickkk
379314
$endgroup$
$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55
add a comment |
$begingroup$
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 '18 at 11:39
mickkkmickkk
379314
$endgroup$
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 '18 at 11:39
mickkkmickkk
379314
asked Nov 22 '18 at 11:39
mickkkmickkk
379314
edited Nov 22 '18 at 11:53
mickkk
asked Nov 22 '18 at 11:39
mickkkmickkk
379314
asked Nov 22 '18 at 11:39
mickkkmickkk
379314
asked Nov 22 '18 at 11:39
mickkkmickkk
379314
379314
$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55
add a comment |
$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55
$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55
$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 15:01
user158565user158565
5,3641518
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add a comment |
$begingroup$
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 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
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 '18 at 15:01
user158565user158565
5,3641518
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:01
user158565user158565
5,3641518
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:01
user158565user158565
5,3641518
$endgroup$
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 '18 at 15:01
user158565user158565
5,3641518
answered Nov 22 '18 at 15:01
user158565user158565
5,3641518
answered Nov 22 '18 at 15:01
user158565user158565
5,3641518
answered Nov 22 '18 at 15:01
user158565user158565
5,3641518
5,3641518
add a comment |
add a comment |
$begingroup$
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 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
$endgroup$
add a comment |
$begingroup$
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 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
$endgroup$
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 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
answered Nov 22 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
answered Nov 22 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
answered Nov 22 '18 at 15:03
Dilip SarwateDilip Sarwate
30k252147
30k252147
add a comment |
add a comment |
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$begingroup$
See stats.stackexchange.com/questions/94872/… inter alia.
$endgroup$
– whuber♦
Nov 22 '18 at 14:55