# Tag Info

## New answers tagged uniform-distribution

0

If your random permutation polynomials are of degree one, they have the form $T(x) = ax+b$ for some $a, b \in \mathbb{Z}_{p}$, $a \neq 0$. I'm not sure how many fixed inputs there are. For a single input $x_{1}$, $T(x_{1}) = ax_{1}+b = y_{1}$ whenever $b = y_{1}-ax_{1}$; there is precisely one $b$ for any choice of $a$, and $p$ choices for $b$, so the ...

3

Clearly, when $b=0$, $v=a$, so it is not a uniform random variable. When $b \neq 0$, we have that $b$ has an inverse in $\mathbb{Z}_p$, so we have for $c \in \mathbb{Z}_p$ $$\mathbb{P} ( v = c) = \mathbb{P} ( a + b r =c) = \mathbb{P} ( b r = c - a) = \mathbb{P}(r =b^{-1}(c-a)) .$$ Since $r$ is uniform, we have $\mathbb{P}( r = \hat{c} ) = \frac{1}{P}$ for ...

1

What I would do is find the area of the space allowed in your problem (depicted) I'd then find the area of the space you're looking to find the probability of landing in, which means adding this inequality: $$t_1 + t_2 > 10$$ Which looks like this: Divide the second area by that of the first, and you get your probability!

1

The pair $(X,Y)$ is uniformly distributed in a triangle. That does not imply that either $X$ or $Y$ is uniformly distributed. Draw the triangle, and you'll see why $X$ is more likely to be in a short interval near $1$ than in an interval of the same length near $0$, so $X$ will not be uniformly distributed. Similarly, $Y$ is more likely to be in a short ...

1

Hint: In general: $$\mathbb E(X_1+\cdots+X_n)=\mathbb EX_1+\cdots+\mathbb EX_n$$ Provided that all expectations $\mathbb EX_i$ exist. Also: $$\mathbb EaX=a\mathbb EX$$ Again provided that expectation $\mathbb EX$ exists. From now on "expectation is linear" should be part of your luggage.

1

Here is a rough and sloppy sketch of a proof. It should not be difficult to fill in any gaps. Let $\{x\}$ refer to the fractional part of $x$. First, we need the following fact $\{a_1+a_2\} = \{a_1\} + \{a_2\} - \lfloor \{a_1\}+\{a_2\} \rfloor$. Let us start with $\{\sum_{i=1}^n U_i\}$. The $U_i$'s are i.i.d uniformly distributed in $[0,1]$. ...

2

A simple argument by induction will also work. If we define $\{x\} = x - \lfloor x \rfloor$ as the fractional part of $x$, then proving that $\{U_1 + U_2\} \sim U \sim \operatorname{Uniform}(0,1)$ will then allow you to inductively show the main result, because $$\{U_1 + \cdots + U_n\} = \{\{U_1 + \cdots + U_{n-1} \} + U_n \}$$ and each $U_1, \ldots, U_n$ ...

1

The characteristic function of the sum ($S_n$) of i.i.d. $U[0,1]$ random variables $$\varphi_{S_n}(2\pi h)=\left(\frac{e^{2\pi h i}-1}{2\pi h i}\right)^n=0$$ for any $h\in\mathbb{Z}\text{\\} \{0\}$. Here is some reference.

4

Note that the vector $(X_1,X_2,...,X_n)$ has density $f(x_1,x_2,...,x_n)=1$ on the unit hypercube. So $\mathbf{P}(S_n\leq 1)$ is just the volume of the region bounded by the axis planes and the plane $x_1+x_2+\cdots+x_n=1$ and equals $1/{n!}$.

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