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 $J<N$ and
$0<q<1$ 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?

Thanks in advance

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

Prove that $grad(f(u(x_1, …, x_m), v(x_1, …, x_m))) = frac{df}{du} gradu + frac{df}{dv} gradv$

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.
enter image description here