Convolution of probabilities on finite groups
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 at 11:40
A.E
62
add a comment |
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 at 11:40
A.E
62
What don't you understand?
– Hempelicious
Dec 8 at 0:45
@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
– A.E
Dec 10 at 10:54
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18
add a comment |
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 at 11:40
A.E
62
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 at 11:40
A.E
62
asked Dec 6 at 11:40
A.E
62
edited Dec 6 at 12:07
asked Dec 6 at 11:40
A.E
62
asked Dec 6 at 11:40
A.E
62
asked Dec 6 at 11:40
A.E
62
62
What don't you understand?
– Hempelicious
Dec 8 at 0:45
@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
– A.E
Dec 10 at 10:54
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18
add a comment |
What don't you understand?
– Hempelicious
Dec 8 at 0:45
@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
– A.E
Dec 10 at 10:54
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18
What don't you understand?
– Hempelicious
Dec 8 at 0:45
What don't you understand?
– Hempelicious
Dec 8 at 0:45
@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
– A.E
Dec 10 at 10:54
@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
– A.E
Dec 10 at 10:54
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18
add a comment |
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What don't you understand?
– Hempelicious
Dec 8 at 0:45
@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
– A.E
Dec 10 at 10:54
It's just the definition of convolution. It says "sum over all pairs of elements in $G$ whose product is $s$".
– Hempelicious
Dec 10 at 13:18