Stack Exchange
log in

Siong Thye Goh

 Discussion between Siong Thye Goh and…

Imported from a comment discussion on math.stackexchange.com/q...
Siong Thye Goh
May 3, 2024 14:40
I don't have a solution for that yet. but I think of another information that perhaps we can use.

if x resides in two orthant and it is optimal in orthant 1, suppose we find a better solution in orthant 2, then x is not a local minima
Siong Thye Goh
May 3, 2024 14:39
for example orthant 1 is x1>=0, x2>=0, x3>=0
orthant 2 is x1 <= 0, x2>=0, x3>=0

Suppose the optimal solution in orthant 1 is (0,0,3)
note that this point also resides in orthant 2. Now my question is is it also optimal in orthant 2?
Siong Thye Goh
May 3, 2024 13:23
I think a mystery that I haven't figure out is if $x$ is in two orthant, $O_1$ and $O_2$, and if $x$ is the optimal solution in $O_1$, is it optimal in $O_2$ as well?
Siong Thye Goh
May 3, 2024 09:46
my initial thought of using those conditions is to comb through 2^N region and use the condition that it is indeed a stationary point. I don't have a proof yet but I wonder if all solution found is a local minimum.
Siong Thye Goh
May 3, 2024 08:25
if we go through each orthant, we would have found all the candidates right?
Siong Thye Goh
May 3, 2024 08:24
true, that is a good point.
Siong Thye Goh
May 3, 2024 08:24
I think it is similar to the concept of stationary point. convexity is not required.
Siong Thye Goh
May 3, 2024 08:24
ah, I misinterpreted your question as you are trying to find all global minimum. Hmm.... thanks for raising the good point. I haven't figured that part as well.
 

 Discussion between Siong Thye Goh and…

Imported from a comment discussion on math.stackexchange.com/q...
Siong Thye Goh
Jan 31, 2022 16:02
E[Y] is a number
Siong Thye Goh
Jan 31, 2022 04:47
welcoem to the site, don't be discouraged from the negative votes :) see you around the site sometimes
Siong Thye Goh
Jan 31, 2022 04:44
welcome. remember for future post, always include your attempt and use descriptive title. :)
Siong Thye Goh
Jan 31, 2022 04:43
seems to be so
Siong Thye Goh
Jan 31, 2022 04:41
yup
Siong Thye Goh
Jan 31, 2022 04:39
$77^2$
Siong Thye Goh
Jan 31, 2022 04:37
the formula for the moments can be found on the wikipedia page
Siong Thye Goh
Jan 31, 2022 04:37
the formula is $3\tilde{\sigma}^4$
Siong Thye Goh
Jan 31, 2022 04:35
we can't just power 4 each individual term
Siong Thye Goh
Jan 31, 2022 04:33
how do you obtain that number?
Siong Thye Goh
Jan 31, 2022 04:31
just starting the chat to avoid long comment threads
Siong Thye Goh
Jan 31, 2022 04:31
in general, square of the sum is not equal to the sum of square of each term
Siong Thye Goh
Jan 31, 2022 04:30
hi, since $\tilde{\sigma}^2 = 77$, we have $\tilde{\sigma}^4 = 77^2$.
Siong Thye Goh
Jan 31, 2022 04:29
yup. $E[X_1^2]=Var(X_1)+E[X_1]^2$.
Siong Thye Goh
Jan 31, 2022 04:29
$E[Y]=E([(X_1+2X_2+X_3)^2]$ this is equal to the second moment of $X_1+2X_2+X_3$ which is just equal to $2^2+8^2+3^2=77$ since the expectation of $X_1+2X_2+X_3$ is $0$.
Siong Thye Goh
Jan 31, 2022 04:29
No. Independence just means $E[X_1X_2]=E[X_1]E[X_2]$.
 

 Doubt old

Old doubt
Siong Thye Goh
Nov 29, 2021 08:29
or if i have misunderstood him
Siong Thye Goh
Nov 29, 2021 08:29
I am not sure if what he claims is true for now
Siong Thye Goh
Nov 29, 2021 08:27
oh well, it is beyond me.
Siong Thye Goh
Nov 29, 2021 08:27
such as $n$ is prime?
Siong Thye Goh
Nov 29, 2021 08:27
did he claim any condition?
Siong Thye Goh
Nov 29, 2021 08:22
i didn't think through carefully but i believed he gave a proof, please check his argument
Siong Thye Goh
Nov 29, 2021 08:21
but i m having a universal statement on one side, so to verify it, you have to check all x
Siong Thye Goh
Nov 29, 2021 08:20
you can try with particular numbers to check if both take the same truth value
Siong Thye Goh
Nov 29, 2021 08:20
are you able to undersrtand the equivalence?
Siong Thye Goh
Nov 29, 2021 08:20
$a^n \equiv 1 \pmod{n} \iff \forall x, a^x \equiv a^r \pmod{n}$.
Siong Thye Goh
Nov 29, 2021 08:19
...
Siong Thye Goh
Nov 29, 2021 08:19
i am claiming a general statement
Siong Thye Goh
Nov 29, 2021 08:10
hence $a^n \equiv 1$ is the sufficient and necessary condition for $a^x \equiv a^r \pmod{n}$.
Siong Thye Goh
Nov 29, 2021 08:10
hence the only way for all $x$, such that $a^x \equiv a^r \pmod{n}$ to hold true is $a^n \equiv 1$.
Siong Thye Goh
Nov 29, 2021 08:09
Now if $a^n \not\equiv 1$, we proved that $a^x \equiv a^r \pmod{n}$ is not true for all $x$.
Siong Thye Goh
Nov 29, 2021 08:09
If $a^n \equiv 1$, we proved that $a^x \equiv a^r \pmod{n}$. Hence $a^n \equiv 1$ is a sufficient condition for $a^x \equiv a^r \pmod{n}$.
Siong Thye Goh
Nov 29, 2021 07:47
that is it is an equivalence statement
Siong Thye Goh
Nov 29, 2021 07:46
and then after which he check $a^n \not \equiv 1$ to check that $a^n \equiv 1$ is also a necessary condition
Siong Thye Goh
Nov 29, 2021 07:46
"if $a^n \equiv 1$, that would be the sufficient condition for $a^x \equiv a^r \pmod{n}$.
Siong Thye Goh
Nov 29, 2021 07:45
he considers $a^n \equiv 1$ and $a^n \not \equiv 1$ separately right
Siong Thye Goh
Nov 29, 2021 07:39
since he considers two cases, he has found the sufficient and necessary condition for all $x$, $a^x \equiv a^r \pmod{n}$.
Siong Thye Goh
Nov 29, 2021 07:39
hence he covers both cases
Siong Thye Goh
Nov 29, 2021 07:38
I think he consider $a^n \equiv 1 \pmod{n}$ and $a^n \not \equiv 1 \pmod{n}$, two cases right?
Siong Thye Goh
Nov 29, 2021 07:38
$a^n \equiv 1 \pmod{n}$ if and only if for all $x,a^x≡a^r\pmod{n}$
Siong Thye Goh
Nov 29, 2021 07:33
i suspect what i stated is what he claimed
Siong Thye Goh
Nov 29, 2021 07:25
that is why i process it and include a quantifier statement