Independence of 4 random variables











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$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










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    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07

















up vote
0
down vote

favorite












$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










share|cite|improve this question


















  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07















up vote
0
down vote

favorite









up vote
0
down vote

favorite











$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!










share|cite|improve this question













$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with



$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.



Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.



Thanks for any help and thoughts!







probability probability-theory independence






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asked Nov 15 at 12:05









DrXYZ

1




1








  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07
















  • 2




    You probably mean events and not random variables. In that case your thinking is correct.
    – Kavi Rama Murthy
    Nov 15 at 12:07










2




2




You probably mean events and not random variables. In that case your thinking is correct.
– Kavi Rama Murthy
Nov 15 at 12:07






You probably mean events and not random variables. In that case your thinking is correct.
– Kavi Rama Murthy
Nov 15 at 12:07












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As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






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    As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



    Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



    Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



    But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



    eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



    If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



    So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



      Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



      Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



      But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



      eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



      If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



      So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






      share|cite|improve this answer























        up vote
        0
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        up vote
        0
        down vote









        As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



        Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



        Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



        But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



        eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



        If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



        So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.






        share|cite|improve this answer












        As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.



        Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...



        Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables ${A_1,A_2}$ and ${B_1,B_2}$.



        But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.



        eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$



        If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.



        So if all those 16 equations are true, you have independence of ${A_1,A_2}$ and ${B_1,B_2}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.







        share|cite|improve this answer












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        answered Nov 15 at 14:04









        Arnaud Mégret

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