Mixing
Proof-verification. If some sets are independent, their complements are independent too
$begingroup$
Could someone verify my proofs?
Suppose (3.2.1) is satisfied.
(3.2.1) $P(A_{alpha_1} cap A_{alpha_2} cap ... A_{alpha_j} ) =
P(A_{alpha_1}) P(A_{alpha_2}) ... P(A_{alpha_j})$ for any $j in mathbb{N}$.
(a) Show that (3.2.1) is still satisfied if $A_{alpha_1}$ is replaced by $A_{alpha_1}^c$.
(b) Show that (3.2.1) is still satisfied if each $A_{alpha_i}$ is replaced
by the corresponding $A_{alpha_i}^c$.
(c) Prove that if ${A_{alpha}}_{alpha in I}$
is independent, then so is
${A_{alpha}}^mathsf{c}_{alpha in I}$.
textbf{Solution.}
(a)
Let $A_{alpha} = bigcaplimits_{k=2}^j A_{alpha_k}$ for any $j geq 2 in mathbb{N}$.
We have to prove that $A_{alpha_1}^mathsf{c}$ is independent of any
$A_{alpha}$.
$P(A_{alpha}) = \
P( (A_{alpha_1} cap A_{alpha})
cup (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) -
P((A_{alpha_1} cap A_{alpha})
cap (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) - P(emptyset)$
Algebrically substituting both sides and using
our assumption que $A_{alpha_1}$ and $A_{alpha}$ are
independents, we have:
$P(A_{alpha_1}^mathsf{c} cap A_{alpha}) = \
P(A_{alpha}) - P(A_{alpha})*P(A_{alpha_1}) = \
P(A_{alpha})*(1 - P(A_{alpha_1})) = \
P(A_{alpha})P(A_{alpha_1}^mathsf{c})$
That's the definition of independence. Then, if $A_{alpha_1}$ is independent of
a any set, $A_{alpha_1}^mathsf{c}$ will have the same property.
(b) We prove by induction. Firstly, we prove the case for $j = 2$.
In this case, if $A_{alpha_1}$ is independent of $A_{alpha_2}$, then
$A_{alpha_1}^mathsf{c}$ is also independent of $A_{alpha_2}$. With
a similar argument, it is easy to check that $A_{alpha_1}^mathsf{c}$ and
$A_{alpha_2}^mathsf{c}$ are also independents.
Assume that (3.2.1) is true for $j leq n$ and that
the expression is also true if we substitute all sets for its complements.
Consider $j = n + 1$. By our assumption, $A_{alpha_1}^mathsf{c},
A_{alpha_2}^mathsf{c}, ..., A_{alpha_n}^mathsf{c}$ are independent.
Let's call their intersecion $A_{alpha}^mathsf{c}$.
By the proposition proved in (a), $A_{alpha}^mathsf{c}$ and
$A_{alpha_{n+1}}$ are independent. For a similar argument,
$A_{alpha}^mathsf{c}$ and $A_{alpha_{n+1}}^mathsf{c}$ are also independent.
$blacksquare$
(c) This is a direct consequence of (b).
probability-theory
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
$endgroup$
add a comment |
$begingroup$
Could someone verify my proofs?
Suppose (3.2.1) is satisfied.
(3.2.1) $P(A_{alpha_1} cap A_{alpha_2} cap ... A_{alpha_j} ) =
P(A_{alpha_1}) P(A_{alpha_2}) ... P(A_{alpha_j})$ for any $j in mathbb{N}$.
(a) Show that (3.2.1) is still satisfied if $A_{alpha_1}$ is replaced by $A_{alpha_1}^c$.
(b) Show that (3.2.1) is still satisfied if each $A_{alpha_i}$ is replaced
by the corresponding $A_{alpha_i}^c$.
(c) Prove that if ${A_{alpha}}_{alpha in I}$
is independent, then so is
${A_{alpha}}^mathsf{c}_{alpha in I}$.
textbf{Solution.}
(a)
Let $A_{alpha} = bigcaplimits_{k=2}^j A_{alpha_k}$ for any $j geq 2 in mathbb{N}$.
We have to prove that $A_{alpha_1}^mathsf{c}$ is independent of any
$A_{alpha}$.
$P(A_{alpha}) = \
P( (A_{alpha_1} cap A_{alpha})
cup (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) -
P((A_{alpha_1} cap A_{alpha})
cap (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) - P(emptyset)$
Algebrically substituting both sides and using
our assumption que $A_{alpha_1}$ and $A_{alpha}$ are
independents, we have:
$P(A_{alpha_1}^mathsf{c} cap A_{alpha}) = \
P(A_{alpha}) - P(A_{alpha})*P(A_{alpha_1}) = \
P(A_{alpha})*(1 - P(A_{alpha_1})) = \
P(A_{alpha})P(A_{alpha_1}^mathsf{c})$
That's the definition of independence. Then, if $A_{alpha_1}$ is independent of
a any set, $A_{alpha_1}^mathsf{c}$ will have the same property.
(b) We prove by induction. Firstly, we prove the case for $j = 2$.
In this case, if $A_{alpha_1}$ is independent of $A_{alpha_2}$, then
$A_{alpha_1}^mathsf{c}$ is also independent of $A_{alpha_2}$. With
a similar argument, it is easy to check that $A_{alpha_1}^mathsf{c}$ and
$A_{alpha_2}^mathsf{c}$ are also independents.
Assume that (3.2.1) is true for $j leq n$ and that
the expression is also true if we substitute all sets for its complements.
Consider $j = n + 1$. By our assumption, $A_{alpha_1}^mathsf{c},
A_{alpha_2}^mathsf{c}, ..., A_{alpha_n}^mathsf{c}$ are independent.
Let's call their intersecion $A_{alpha}^mathsf{c}$.
By the proposition proved in (a), $A_{alpha}^mathsf{c}$ and
$A_{alpha_{n+1}}$ are independent. For a similar argument,
$A_{alpha}^mathsf{c}$ and $A_{alpha_{n+1}}^mathsf{c}$ are also independent.
$blacksquare$
(c) This is a direct consequence of (b).
probability-theory
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
$endgroup$
1
$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19
add a comment |
$begingroup$
Could someone verify my proofs?
Suppose (3.2.1) is satisfied.
(3.2.1) $P(A_{alpha_1} cap A_{alpha_2} cap ... A_{alpha_j} ) =
P(A_{alpha_1}) P(A_{alpha_2}) ... P(A_{alpha_j})$ for any $j in mathbb{N}$.
(a) Show that (3.2.1) is still satisfied if $A_{alpha_1}$ is replaced by $A_{alpha_1}^c$.
(b) Show that (3.2.1) is still satisfied if each $A_{alpha_i}$ is replaced
by the corresponding $A_{alpha_i}^c$.
(c) Prove that if ${A_{alpha}}_{alpha in I}$
is independent, then so is
${A_{alpha}}^mathsf{c}_{alpha in I}$.
textbf{Solution.}
(a)
Let $A_{alpha} = bigcaplimits_{k=2}^j A_{alpha_k}$ for any $j geq 2 in mathbb{N}$.
We have to prove that $A_{alpha_1}^mathsf{c}$ is independent of any
$A_{alpha}$.
$P(A_{alpha}) = \
P( (A_{alpha_1} cap A_{alpha})
cup (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) -
P((A_{alpha_1} cap A_{alpha})
cap (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) - P(emptyset)$
Algebrically substituting both sides and using
our assumption que $A_{alpha_1}$ and $A_{alpha}$ are
independents, we have:
$P(A_{alpha_1}^mathsf{c} cap A_{alpha}) = \
P(A_{alpha}) - P(A_{alpha})*P(A_{alpha_1}) = \
P(A_{alpha})*(1 - P(A_{alpha_1})) = \
P(A_{alpha})P(A_{alpha_1}^mathsf{c})$
That's the definition of independence. Then, if $A_{alpha_1}$ is independent of
a any set, $A_{alpha_1}^mathsf{c}$ will have the same property.
(b) We prove by induction. Firstly, we prove the case for $j = 2$.
In this case, if $A_{alpha_1}$ is independent of $A_{alpha_2}$, then
$A_{alpha_1}^mathsf{c}$ is also independent of $A_{alpha_2}$. With
a similar argument, it is easy to check that $A_{alpha_1}^mathsf{c}$ and
$A_{alpha_2}^mathsf{c}$ are also independents.
Assume that (3.2.1) is true for $j leq n$ and that
the expression is also true if we substitute all sets for its complements.
Consider $j = n + 1$. By our assumption, $A_{alpha_1}^mathsf{c},
A_{alpha_2}^mathsf{c}, ..., A_{alpha_n}^mathsf{c}$ are independent.
Let's call their intersecion $A_{alpha}^mathsf{c}$.
By the proposition proved in (a), $A_{alpha}^mathsf{c}$ and
$A_{alpha_{n+1}}$ are independent. For a similar argument,
$A_{alpha}^mathsf{c}$ and $A_{alpha_{n+1}}^mathsf{c}$ are also independent.
$blacksquare$
(c) This is a direct consequence of (b).
probability-theory
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
$endgroup$
Could someone verify my proofs?
Suppose (3.2.1) is satisfied.
(3.2.1) $P(A_{alpha_1} cap A_{alpha_2} cap ... A_{alpha_j} ) =
P(A_{alpha_1}) P(A_{alpha_2}) ... P(A_{alpha_j})$ for any $j in mathbb{N}$.
(a) Show that (3.2.1) is still satisfied if $A_{alpha_1}$ is replaced by $A_{alpha_1}^c$.
(b) Show that (3.2.1) is still satisfied if each $A_{alpha_i}$ is replaced
by the corresponding $A_{alpha_i}^c$.
(c) Prove that if ${A_{alpha}}_{alpha in I}$
is independent, then so is
${A_{alpha}}^mathsf{c}_{alpha in I}$.
textbf{Solution.}
(a)
Let $A_{alpha} = bigcaplimits_{k=2}^j A_{alpha_k}$ for any $j geq 2 in mathbb{N}$.
We have to prove that $A_{alpha_1}^mathsf{c}$ is independent of any
$A_{alpha}$.
$P(A_{alpha}) = \
P( (A_{alpha_1} cap A_{alpha})
cup (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) -
P((A_{alpha_1} cap A_{alpha})
cap (A_{alpha_1}^mathsf{c} cap A_{alpha})) = \
P(A_{alpha_1} cap A_{alpha}) +
P(A_{alpha_1}^mathsf{c} cap A_{alpha}) - P(emptyset)$
Algebrically substituting both sides and using
our assumption que $A_{alpha_1}$ and $A_{alpha}$ are
independents, we have:
$P(A_{alpha_1}^mathsf{c} cap A_{alpha}) = \
P(A_{alpha}) - P(A_{alpha})*P(A_{alpha_1}) = \
P(A_{alpha})*(1 - P(A_{alpha_1})) = \
P(A_{alpha})P(A_{alpha_1}^mathsf{c})$
That's the definition of independence. Then, if $A_{alpha_1}$ is independent of
a any set, $A_{alpha_1}^mathsf{c}$ will have the same property.
(b) We prove by induction. Firstly, we prove the case for $j = 2$.
In this case, if $A_{alpha_1}$ is independent of $A_{alpha_2}$, then
$A_{alpha_1}^mathsf{c}$ is also independent of $A_{alpha_2}$. With
a similar argument, it is easy to check that $A_{alpha_1}^mathsf{c}$ and
$A_{alpha_2}^mathsf{c}$ are also independents.
Assume that (3.2.1) is true for $j leq n$ and that
the expression is also true if we substitute all sets for its complements.
Consider $j = n + 1$. By our assumption, $A_{alpha_1}^mathsf{c},
A_{alpha_2}^mathsf{c}, ..., A_{alpha_n}^mathsf{c}$ are independent.
Let's call their intersecion $A_{alpha}^mathsf{c}$.
By the proposition proved in (a), $A_{alpha}^mathsf{c}$ and
$A_{alpha_{n+1}}$ are independent. For a similar argument,
$A_{alpha}^mathsf{c}$ and $A_{alpha_{n+1}}^mathsf{c}$ are also independent.
$blacksquare$
(c) This is a direct consequence of (b).
probability-theory
probability-theory
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
edited Jan 14 at 23:22
DunhoClark
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
asked Jan 14 at 16:56
DunhoClarkDunhoClark
505
505
1
$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19
add a comment |
1
$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19
1
1
$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19
add a comment |
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$begingroup$
For a) the idea is good but you only proved pairwise independence, you should directly do the same with all the $A_{alpha_i}$ sets. It is true that b) is applying the first result several times but be careful again with the difference between pairwise independence and independence. For c) I would say that it is a direct consequence of a), not b). This is because you can conjugate any number of the sets which is the definition of independence.
$endgroup$
– P. Quinton
Jan 14 at 17:28
$begingroup$
@P.Quinton, but, by the associative property of the algebra of sets, the pairwise independence extends to all the sets, right?
$endgroup$
– DunhoClark
Jan 14 at 17:54
$begingroup$
well unless you specify the set $A_{alpha_i}$ you are considering the property you saying is satisfied, then my guess is that the property is just true for those special sets. And then my guess is that pairwise independence implying independence would be a known fact if it were true, there are some counter examples.
$endgroup$
– P. Quinton
Jan 14 at 21:19