Convolution of probabilities on finite groups
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I was reading a book on group and representation theory and came across the following which I don't understand, I'd appreciate any help.
Suppose P and Q are probabilities on a finite group G. Thus $P(s)ge0$ and $sum_s P(s)=1$. By the convolution $P*Q$ we mean the probability $P*Q(s)=sum_t P(st^{-1})Q(t)$; "first pick $t$ from $Q$, then independently pick $u$ from $P$ and form the product $ut.$" Note that in general $P*Q ne Q*P$.
EDIT: the first line is standard, but the latter I cant figure.
group-theory finite-groups representation-theory convolution
asked Dec 6 '18 at 11:40
A.EA.E
285
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add a comment |
$begingroup$
I was reading a book on group and representation theory and came across the following which I don't understand, I'd appreciate any help.
Suppose P and Q are probabilities on a finite group G. Thus $P(s)ge0$ and $sum_s P(s)=1$. By the convolution $P*Q$ we mean the probability $P*Q(s)=sum_t P(st^{-1})Q(t)$; "first pick $t$ from $Q$, then independently pick $u$ from $P$ and form the product $ut.$" Note that in general $P*Q ne Q*P$.
EDIT: the first line is standard, but the latter I cant figure.
group-theory finite-groups representation-theory convolution
asked Dec 6 '18 at 11:40
A.EA.E
285
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$begingroup$
What don't you understand?
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– Hempelicious
Dec 8 '18 at 0:45
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@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
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– A.E
Dec 10 '18 at 10:54
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18
add a comment |
$begingroup$
I was reading a book on group and representation theory and came across the following which I don't understand, I'd appreciate any help.
Suppose P and Q are probabilities on a finite group G. Thus $P(s)ge0$ and $sum_s P(s)=1$. By the convolution $P*Q$ we mean the probability $P*Q(s)=sum_t P(st^{-1})Q(t)$; "first pick $t$ from $Q$, then independently pick $u$ from $P$ and form the product $ut.$" Note that in general $P*Q ne Q*P$.
EDIT: the first line is standard, but the latter I cant figure.
group-theory finite-groups representation-theory convolution
asked Dec 6 '18 at 11:40
A.EA.E
285
$endgroup$
I was reading a book on group and representation theory and came across the following which I don't understand, I'd appreciate any help.
Suppose P and Q are probabilities on a finite group G. Thus $P(s)ge0$ and $sum_s P(s)=1$. By the convolution $P*Q$ we mean the probability $P*Q(s)=sum_t P(st^{-1})Q(t)$; "first pick $t$ from $Q$, then independently pick $u$ from $P$ and form the product $ut.$" Note that in general $P*Q ne Q*P$.
EDIT: the first line is standard, but the latter I cant figure.
group-theory finite-groups representation-theory convolution
group-theory finite-groups representation-theory convolution
asked Dec 6 '18 at 11:40
A.EA.E
285
asked Dec 6 '18 at 11:40
A.EA.E
285
edited Dec 6 '18 at 12:07
A.E
asked Dec 6 '18 at 11:40
A.EA.E
285
asked Dec 6 '18 at 11:40
A.EA.E
285
asked Dec 6 '18 at 11:40
A.EA.E
285
285
$begingroup$
What don't you understand?
$endgroup$
– Hempelicious
Dec 8 '18 at 0:45
$begingroup$
@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
$endgroup$
– A.E
Dec 10 '18 at 10:54
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18
add a comment |
$begingroup$
What don't you understand?
$endgroup$
– Hempelicious
Dec 8 '18 at 0:45
$begingroup$
@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
$endgroup$
– A.E
Dec 10 '18 at 10:54
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18
$begingroup$
What don't you understand?
$endgroup$
– Hempelicious
Dec 8 '18 at 0:45
$begingroup$
What don't you understand?
$endgroup$
– Hempelicious
Dec 8 '18 at 0:45
$begingroup$
@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
$endgroup$
– A.E
Dec 10 '18 at 10:54
$begingroup$
@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
$endgroup$
– A.E
Dec 10 '18 at 10:54
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18
add a comment |
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$begingroup$
What don't you understand?
$endgroup$
– Hempelicious
Dec 8 '18 at 0:45
$begingroup$
@Hempelicious, I don't understand why P*Q(s) is equal to the sum of P(st^(-1))Q(t). where did the "t" come from
$endgroup$
– A.E
Dec 10 '18 at 10:54
$begingroup$
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
$endgroup$
– Hempelicious
Dec 10 '18 at 13:18