## if 𝑋⊥𝑌∣𝑍 and 𝑋⊥𝑍∣𝑌 then Prove or disprove that 𝑋⊥𝑌 and X⊥Z

if X,Y,Z be random variables, and suppose that the joint distribution is positive and 𝑋⊥𝑌∣𝑍 and 𝑋⊥𝑍∣𝑌 then Prove or disprove that 𝑋⊥𝑌 and X⊥Z

## sequences and series – How to prove that these partial binomial sums are zero?

I am trying to prove that the following equation is equal to zero.
$$0= sum_{j=J+1}^N Big(j (1-q)+ (j-J) (q N-j) Big) cdot q^{j} (1-q)^{N -j} binom{N}{j} label{zero1}$$

Where
$$J,N in mathbb{Z}^+$$ and $$J and
$$0 is a probability.

Numerical simulations (see link) suggest that this is true.
Tips and hints (or solutions) are very welcome!

## hash – How can I prove or disprove that the following function is bijection?

For a research project, I tried to prove or disprove that a function called xxhash128_low is a bijection from 64 bit unsigned integer to 64 bit unsigned integer. I have shown that it is sufficient to prove that the following critical code is a bijection:

Input: a 64-bit integer x
0. Let c=0x9E3779B185EBCB87
1. Let x_low= the 8-low bytes of c*x.
2. Let x_high= the 8-high bytes of c*x.
3. w= x_high+2*x_low.
4  y= shift right 3 bits of w.
5. return x_low XOR y.

What I have proved till now: returning only the low 64 bits, x_low, can be shown to be a permutation. Also, y can be expressed as a function of x_low (i.e., y=f(x_low)).

## Differential equation prove – Mathematics Stack Exchange

I’m starting the differential equation course and I would really appreciate if anyone can help me with ideas solving this prove:

Let a, b:$$Ito R$$ continuous functions with $$I subseteq R$$ and $$U: I to R$$ the solution of the differential equation $$u”= a(t)cdot u + b(t)cdot u’$$ with $$u(t_0)=0$$ and $$u'(t_0)=1$$ for some $$t_0$$ belonging to $$I$$. Show that if a>0 then $$u$$ is increasing in $${t in I: tgeq t_0}$$

## real analysis – Method to Prove That Two Sets Are Equal

When it comes to proving that two sets are equal, say $$A = B$$, we’re usually told that we have to prove that $$A subset B$$ and $$B subset A$$. However, I’m under the impression that this strategy isn’t unique. Two sets are equal if they have the same elements, and this commonly used strategy is just one way to prove that.

A better way to prove that two sets are equal, at least in my opinion, is to use set notation. However, because I never see this being used, I’m unsure if this method is correct. Example:

Prove that $$(A cup B) times C = (A times C) cup (B times C)$$.

Proof. We have,
begin{align} (A cup B) times C &= {x; x in A operatorname{or} B } times {y; y in C } \ &= {(x, y); x in A operatorname{or} B, y in C } \ &= {(x, y); x in A, y in C } cup {(x, y); x in B, y in C } \ &= (A times C) cup (B times C). end{align}

So the question is, is this method correct? Thanks in advance.

## given \$L\$, a regular language, prove that \$L^{|2|}={w_1w_2 text{ | } w_1,w_2in L, |w_1|=|w_2|}\$ is context-free

I have found a problem about proving whether $$L^{|2|}={w_1w_2 text{ | } w_1,w_2in L, |w_1|=|w_2|}$$ is context-free or not, knowing that $$L$$ is regular

So far I know that:

• There are examples where $$L$$ is regular and $$L^{|2|}$$ is regular (for example $$L={a,b}$$)
• There are examples where $$L$$ is regular and $$L^{|2|}$$ is not (for example $$L={w| w=a^N text{ or } w= b^N , Nge0}$$)

But I am not sure how to prove that it’s context-free regardless of which regular language I use. I have found similar problems with the same language without imposing restrictions on which words to use, but I am not sure if those apply to this one.

## logic – how to prove equivalence of two substitutions by induction

I’m trying to prove the following reduction
t{x:=u}{y:=v} = t{y:=v}{x:=u{y:=v}}
We have the following Assuming:
(1)($$x neq y$$ )

(2)x is not a free Variable of v (i.e $$x notin$$fv$$(v)$$🙂

My idea is to do it by induction, but I’m a bit stuck with the base case. I would appreciate if someone can explain to me how to such a proof. I’m including what I tried (but it’s wrong/incomplete)

*Base case*: Assuming t is a just a variable `m` (t=m),
#on the left side we have
if m = x ==> u{y:=v}  (if u=y then `v`  else `u`)
else   ==> m{y:=v} (if m=y then `v`  else `m`)

#On the right side
if m=y ==> v{x:=u{y:=v}}  (if u=y then `v{x:=v}`  else `u{x:=u}` ( so we get either (v=x==> `v` else `v`) or else u=x==> `u` or else `u`) )
else   ==> m{x:=u{y:=v}} (if m=y then `m{x:=v}`  else `m{x:=u}`( so we get either v or m))

I understand we end up getting the same four branches, but is that considered really a proof? Is this the proper way to write such proof and conclude for the base case? Also the given assumptions didn’t help much here so I think I’m missing the part where I need to use those assumptions…

I think once the base case is proven,

we can do the following

case t = ($$t_{1}t_{2}$$)

t{x:=u}{y:=v} ==> t1t2{x:=u}{y:=v}
==> t1{x:=u}{y:=v} t2{x:=u}{y:=v} and we have just proven in the base case that:
t1{x:=u}{y:=v} = t1{y:=v}{x:=u{y:=v}}  and  t2{x:=u}{y:=v} = t2{y:=v}{x:=u{y:=v}}
so  t1{x:=u}{y:=v} t2{x:=u}{y:=v} = t1{y:=v}{x:=u{y:=v}} t2{y:=v}{x:=u{y:=v}}
= t1t2{x:=u}{y:=v} = t1t2{y:=v}{x:=u{y:=v}}

Now only part left is if $$t= lambda m . t$$

so we have $$(lambda m . t)$${x:=u}{y:=v}
this can be directly re-written as $$lambda m . t$${x:=u}{y:=v}, which is the base case again…

Could someone please help ne finish this proof correctly and explain to me the right way to do it?

## How could I prove this? The area of a triangle can be expressed in terms of the medians by..

The area of a triangle can be expressed in terms of the medians by…

by formula

enter image description here

I don$$`$$t know how to prove that, help pls:(

## calculus and analysis – How to prove the following integration identity?

I have the following integration that I want to evaluate it using Mathematica.

Evaluate(Sqrt((Theta))/(Sqrt(p ) BesselK(1, Sqrt(p (Theta))))
Integrate(
x BesselK(0,
x Sqrt(p (Theta))) (Sqrt((Theta))/Sqrt(
s ) (rTD BesselK(1,
rTD Sqrt(s (Theta))) (BesselI(0, x Sqrt(s (Theta))) -
BesselK(0, x Sqrt(s (Theta))) BesselI(0, Sqrt(
s (Theta)))/BesselK(0, Sqrt(s (Theta))))
+
BesselI(0,
x Sqrt(s (Theta))) (x BesselK(1, x Sqrt(s (Theta))) -
rTD BesselK(1, rTD Sqrt(s (Theta)))) HeavisideTheta(
x - rTD)
+
BesselK(0,
x Sqrt(s (Theta))) (x BesselI(1, x Sqrt(s (Theta))) -
rTD BesselI(1, rTD Sqrt(s (Theta)))) HeavisideTheta(
x - rTD))), {x, 1, Infinity}))

I find problem in obtaining the solution using Mathematica, so I tried to evaluate it manually.

I want to make sure that my solution is correct. My solution is as follows.