Mixing
Proving $ A_i $ and $ A_j $ are independent events
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0
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I toss a fair coin $ 10 $ times. $ A_i (i = 1,2,dots,9) $ is the event of the same result in the $ i $ and $ i + 1 $ toss. How do I prove that for every $ i not = j $ the events $ A_i, A_j $ are independent?
I understand that in order to prove the independent of those two events I need to prove that $$ P(A_i | A_j) = P(A_i) P(A_j) $$
I guess I can conut on the fact that $ i, i+1 $ are indepedent events because it is a fair coin. so $ P(A_i) = P(A_j) = frac{1}{2} $ because I can have whatever result I want in the first toss and then I need the same results in the second. so
$$ P(A_j|A_i) = frac{P(A_jcap A_i)}{P(A_i)} = frac{frac{1}{2}^3}{frac{1}{2}} = frac{1}{4} = P(A_i)P(A_j) $$
But what if for example $ i =1 , j = 2 $ it means I should have the same result only in the first, second and third toss and then $ P(A_j | A_i) not = P(A_i)P(A_j) $
am I correct?
probability conditional-probability
asked 9 hours ago
bm1125
54916
add a comment |
up vote
0
down vote
favorite
I toss a fair coin $ 10 $ times. $ A_i (i = 1,2,dots,9) $ is the event of the same result in the $ i $ and $ i + 1 $ toss. How do I prove that for every $ i not = j $ the events $ A_i, A_j $ are independent?
I understand that in order to prove the independent of those two events I need to prove that $$ P(A_i | A_j) = P(A_i) P(A_j) $$
I guess I can conut on the fact that $ i, i+1 $ are indepedent events because it is a fair coin. so $ P(A_i) = P(A_j) = frac{1}{2} $ because I can have whatever result I want in the first toss and then I need the same results in the second. so
$$ P(A_j|A_i) = frac{P(A_jcap A_i)}{P(A_i)} = frac{frac{1}{2}^3}{frac{1}{2}} = frac{1}{4} = P(A_i)P(A_j) $$
But what if for example $ i =1 , j = 2 $ it means I should have the same result only in the first, second and third toss and then $ P(A_j | A_i) not = P(A_i)P(A_j) $
am I correct?
probability conditional-probability
asked 9 hours ago
bm1125
54916
you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I toss a fair coin $ 10 $ times. $ A_i (i = 1,2,dots,9) $ is the event of the same result in the $ i $ and $ i + 1 $ toss. How do I prove that for every $ i not = j $ the events $ A_i, A_j $ are independent?
I understand that in order to prove the independent of those two events I need to prove that $$ P(A_i | A_j) = P(A_i) P(A_j) $$
I guess I can conut on the fact that $ i, i+1 $ are indepedent events because it is a fair coin. so $ P(A_i) = P(A_j) = frac{1}{2} $ because I can have whatever result I want in the first toss and then I need the same results in the second. so
$$ P(A_j|A_i) = frac{P(A_jcap A_i)}{P(A_i)} = frac{frac{1}{2}^3}{frac{1}{2}} = frac{1}{4} = P(A_i)P(A_j) $$
But what if for example $ i =1 , j = 2 $ it means I should have the same result only in the first, second and third toss and then $ P(A_j | A_i) not = P(A_i)P(A_j) $
am I correct?
probability conditional-probability
asked 9 hours ago
bm1125
54916
I toss a fair coin $ 10 $ times. $ A_i (i = 1,2,dots,9) $ is the event of the same result in the $ i $ and $ i + 1 $ toss. How do I prove that for every $ i not = j $ the events $ A_i, A_j $ are independent?
I understand that in order to prove the independent of those two events I need to prove that $$ P(A_i | A_j) = P(A_i) P(A_j) $$
I guess I can conut on the fact that $ i, i+1 $ are indepedent events because it is a fair coin. so $ P(A_i) = P(A_j) = frac{1}{2} $ because I can have whatever result I want in the first toss and then I need the same results in the second. so
$$ P(A_j|A_i) = frac{P(A_jcap A_i)}{P(A_i)} = frac{frac{1}{2}^3}{frac{1}{2}} = frac{1}{4} = P(A_i)P(A_j) $$
But what if for example $ i =1 , j = 2 $ it means I should have the same result only in the first, second and third toss and then $ P(A_j | A_i) not = P(A_i)P(A_j) $
am I correct?
probability conditional-probability
probability conditional-probability
asked 9 hours ago
bm1125
54916
asked 9 hours ago
bm1125
54916
asked 9 hours ago
bm1125
54916
asked 9 hours ago
bm1125
54916
asked 9 hours ago
bm1125
54916
54916
you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago
add a comment |
you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago
you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago
you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
add a comment |
up vote
1
down vote
Two events are independent if
begin{equation}
P(Acap B) = P(A)P(B).
end{equation}
Alterativelly if
begin{equation}
P(A| B) = P(A).
end{equation}
Also $P(A_j cap A_i) =frac{1}{4} iff i+1=j.
answered 9 hours ago
vermator
464
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
add a comment |
up vote
1
down vote
You are asked to prove that for $A_1,dots,A_{10}$ are pairwise independent.
In this answer I go for a stronger result: $A_1,dots,A_{10}$ are mutually independent.
See here for the difference.
Work in probability space $(Omega,wp(Omega),P)$ where: $$Omega={H,T}^{10}={(omega_1,dots,omega_{10})mid omega_iin{H,T}text{ for every }iin{1,dots,10}}$$and $P({omega})=2^{-10}$ for every $omegainOmega$.
Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.
Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.
So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$
This allows the conclusion that $A_1,dots,A_9$ are mutually independent.
Consequently $A_1,dots,A_9$ are also pairwise independent.
answered 9 hours ago
drhab
94.2k543125
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
add a comment |
up vote
1
down vote
accepted
$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
$P(A_1cap A_2)=frac 1 8 +frac 1 8=frac 1 4$ and $P(A_1)=P(A_2)=frac 1 2$ so $A_1$ and $A_2$ are independent.
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
answered 9 hours ago
Kavi Rama Murthy
39.9k31750
39.9k31750
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
add a comment |
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
Hey thanks for your answer. Can you explain just why $ P(A_1 cap A_2) = frac{1}{8} + frac{1}{8} $ I'm not sure what am I missing..
– bm1125
9 hours ago
1
1
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
Probability of all three heads is $frac 1 8$ and probability of all three tails is also $frac 1 8$. You have to add these to get probability of $A_1cap A_2$.
– Kavi Rama Murthy
9 hours ago
add a comment |
up vote
1
down vote
Two events are independent if
begin{equation}
P(Acap B) = P(A)P(B).
end{equation}
Alterativelly if
begin{equation}
P(A| B) = P(A).
end{equation}
Also $P(A_j cap A_i) =frac{1}{4} iff i+1=j.
answered 9 hours ago
vermator
464
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
add a comment |
up vote
1
down vote
Two events are independent if
begin{equation}
P(Acap B) = P(A)P(B).
end{equation}
Alterativelly if
begin{equation}
P(A| B) = P(A).
end{equation}
Also $P(A_j cap A_i) =frac{1}{4} iff i+1=j.
answered 9 hours ago
vermator
464
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Two events are independent if
begin{equation}
P(Acap B) = P(A)P(B).
end{equation}
Alterativelly if
begin{equation}
P(A| B) = P(A).
end{equation}
Also $P(A_j cap A_i) =frac{1}{4} iff i+1=j.
answered 9 hours ago
vermator
464
Two events are independent if
begin{equation}
P(Acap B) = P(A)P(B).
end{equation}
Alterativelly if
begin{equation}
P(A| B) = P(A).
end{equation}
Also $P(A_j cap A_i) =frac{1}{4} iff i+1=j.
answered 9 hours ago
vermator
464
edited 9 hours ago
answered 9 hours ago
vermator
464
answered 9 hours ago
vermator
464
answered 9 hours ago
vermator
464
464
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
add a comment |
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
Thanks for the answer, Let me make sure I understand $ P(A_j cap A_i) $ is the probability of having the same result in four tosses? Shouldn't it be $ frac{1}{2}^3 $ then? Because I can have whatever result on the first throw and then the next three throws should have the same result. What am I missing?
– bm1125
9 hours ago
1
1
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
I thought that $i+1 = j$, my mistake :) In general it's $frac{1}{8}$, but when $j=i+1$ then it's $frac{1}{4}$
– vermator
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
No, you are right. I specifically asked about the case of $ i+1 = j $ . Thanks for clarifying!
– bm1125
9 hours ago
add a comment |
up vote
1
down vote
You are asked to prove that for $A_1,dots,A_{10}$ are pairwise independent.
In this answer I go for a stronger result: $A_1,dots,A_{10}$ are mutually independent.
See here for the difference.
Work in probability space $(Omega,wp(Omega),P)$ where: $$Omega={H,T}^{10}={(omega_1,dots,omega_{10})mid omega_iin{H,T}text{ for every }iin{1,dots,10}}$$and $P({omega})=2^{-10}$ for every $omegainOmega$.
Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.
Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.
So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$
This allows the conclusion that $A_1,dots,A_9$ are mutually independent.
Consequently $A_1,dots,A_9$ are also pairwise independent.
answered 9 hours ago
drhab
94.2k543125
add a comment |
up vote
1
down vote
You are asked to prove that for $A_1,dots,A_{10}$ are pairwise independent.
In this answer I go for a stronger result: $A_1,dots,A_{10}$ are mutually independent.
See here for the difference.
Work in probability space $(Omega,wp(Omega),P)$ where: $$Omega={H,T}^{10}={(omega_1,dots,omega_{10})mid omega_iin{H,T}text{ for every }iin{1,dots,10}}$$and $P({omega})=2^{-10}$ for every $omegainOmega$.
Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.
Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.
So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$
This allows the conclusion that $A_1,dots,A_9$ are mutually independent.
Consequently $A_1,dots,A_9$ are also pairwise independent.
answered 9 hours ago
drhab
94.2k543125
add a comment |
up vote
1
down vote
up vote
1
down vote
You are asked to prove that for $A_1,dots,A_{10}$ are pairwise independent.
In this answer I go for a stronger result: $A_1,dots,A_{10}$ are mutually independent.
See here for the difference.
Work in probability space $(Omega,wp(Omega),P)$ where: $$Omega={H,T}^{10}={(omega_1,dots,omega_{10})mid omega_iin{H,T}text{ for every }iin{1,dots,10}}$$and $P({omega})=2^{-10}$ for every $omegainOmega$.
Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.
Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.
So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$
This allows the conclusion that $A_1,dots,A_9$ are mutually independent.
Consequently $A_1,dots,A_9$ are also pairwise independent.
answered 9 hours ago
drhab
94.2k543125
You are asked to prove that for $A_1,dots,A_{10}$ are pairwise independent.
In this answer I go for a stronger result: $A_1,dots,A_{10}$ are mutually independent.
See here for the difference.
Work in probability space $(Omega,wp(Omega),P)$ where: $$Omega={H,T}^{10}={(omega_1,dots,omega_{10})mid omega_iin{H,T}text{ for every }iin{1,dots,10}}$$and $P({omega})=2^{-10}$ for every $omegainOmega$.
Let $E_iin{A_i,A_i^{complement}}$ for $i=1,dots,9$.
Every $omega=(omega_1,dots,omega_{10})in E_1capcdotscap E_9$ is completely determined by $omega_1$ so that $E_1capcdotscap E_9$ contains exactly $2$ elements of $Omega$.
So $P(E_1capcdotscap E_9)=2cdot2^{-10}=2^{-9}=P(E_1)timescdotstimes P(E_{9})$
This allows the conclusion that $A_1,dots,A_9$ are mutually independent.
Consequently $A_1,dots,A_9$ are also pairwise independent.
answered 9 hours ago
drhab
94.2k543125
edited 8 hours ago
answered 9 hours ago
drhab
94.2k543125
answered 9 hours ago
drhab
94.2k543125
answered 9 hours ago
drhab
94.2k543125
94.2k543125
add a comment |
add a comment |
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you nts that $mathsf{P}(A_icap A_j)=mathsf{P}(A_i)mathsf{P}(A_j)$ or $mathsf{P}(A_imid A_j)=mathsf{P}(A_i)$...
– d.k.o.
9 hours ago