If $d_i \in U(-1,1)$ (uniform distribution between -1 and 1 - not sure what the canonical notation is for this), then it seems intuitive that $\sum_{i=1}^n d_i \in U(-n,n)$ and thus $$P\big(\sum_{i=1}^n d_i &gt; 0\big) = \frac{1}{2}$$.  If this is true, it also seems intuitive that $P\big(\sum_{i=1}^n p_id_i\big) \in U(-n,n)$ where $p_i \in {-1,1}$. That is, we can apply the same ideas to a combination of additions/subtractions, not just a summation.  I'm not classically trained in statistics and don't know if this is a standard theorem or if I am way off base here. Is this true / false? If it is true, is it provable using classical statistical techniques?
22342
I've been wondering about this one for a while; I find it a little weird how abruptly it happens. Basically, why do we need just three uniforms for $Z_n$ to smooth out like it does? And why does the smoothing-out happen so relatively quickly?  $Z_2$:     $Z_3$:    (images shamelessly stolen from John D. Cook's blog: )  Why doesn't it take, say, four uniforms? Or five? Or...?
17814
It is easily checked that $\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem.  I'm trying to figure out what happens with the truncated sum $\displaystyle\sum_{i\ =\ 0}^{D}\left(\, -1\,\right)^{i}\binom{n}{i}$.  How far away from $0$ can this get, as a function of $D$ ?.  I'm mostly interested in the case of when $D \ll n$, such as $D \sim \,\sqrt{\,n\,}\,$.  Thanks !
120911
We have that $N$ and $X_1, X_2, \dots$ are all independent. We also have $\operatorname{E} [X_j] = \mu$ and $\operatorname{Var}[X_j] = σ^2$.  Then, we introduce an integer–valued random variable, $N$, which is the random sum such that: $$Z = \sum_{j=1}^{N+1}X_j.$$ Assuming that $N$ is distributed $\sim\mathrm{Poisson}(\lambda)$, what is the first moment and what is the variance of $Z$?  For a normal Poisson distribution, I know the variance is just $\lambda$, as is the mean. I'm having trouble understanding the implication of having the bounds be poisson distributed. Normally, I would just say "variance of the sum is the sum of the variance," but I don't think that's how it works with random sums. Any hints/guidance appreciated.
122651
Is $\sum_{n=2}^\infty (n\ln n)^{-1}=\infty$ ?  This seems like elementary calculus, but I can't figure this out.  Can anyone supply a hint?
76207
Can someone please explain to me how is $$\dfrac1{(1-x)^5}=\sum_{n\geq0}{n+4\choose4}x^n$$  Thanks!
97524
Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!
9136
I am a Biology major (please don't shame me!) but I really enjoy mathematics. Recently I have been reading about this conjecture and its importance in understanding the distribution of primes. After being studied for so long, how come all attempts have failed? Why is this such a difficult thing to study and work on? Would finding a closed form of $$\sum_1^\infty \frac {1}{n^s}$$ help the issue? Have any attempts been made at this at all? Thank you very much for your time educating me and your help!!
88502
Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
121714
$X_{n}$ independent and $X_n \sim \mathcal{P}(n) $  meaning that $X_{n}$ has Poisson distributions with parameter $n$. What is the $\lim\limits_{n\to \infty} \frac{X_{n}}{n}$ almost surely ?    I think we can write $X(n) \sim X(1)+X(1)+\cdots+X(1)$ where the sum is taken on independent identical distribution then use the law of large number. But I am not sure that is it correct or not. Can anyone give me some hints? Thank you in advance!
64072
Let $X_1(x),X_2(x),\ldots ,X_n(x)$ be $n$ iid random variables. I know that  $$\mathbf{P}(\lim_n X_n(x)=X(x))=1$$ for each $x\in \mathbb{R}$. We also know that the limit random variable $X$ is a.s. constant and considering $X\colon \mathbb{R}\to\mathbb{R}$, it is continuous. I want to show that  $$\mathbf{P}(\lim_nX_n(x)=X(x)\ \text{ for all }x\in \mathbb{R})=1.$$ I thought about using countable additivity of probability on $\mathbb{Q}$ and then applying continuouty to extend the result to the whole $\mathbb{R}$. Since $$\mathbf{P}(\lim_nX_n(x)\neq X(x)\ \text{ for all }x\in \mathbb{Q})=\sum_{x\in \mathbb{Q}}\mathbb{P}(X_n(x)\neq X(x))=0+0+\ldots.$$ Is there anything else I should add to this explanation? Ideas?
38315
I am working on the following problem:     Let $X_1, X_2, \ldots, X_n, \ldots$ be iid. random variables, each of Uniform$(0,1)$ distribution. Denote by $f_n(x)$ the density of the random variable $S_n := \sum_{k = 1}^n X_k$. Then    \begin{align*} f_n(x) = \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{n}{k} (x-k)^{n-1}, \end{align*}   where $[x]$ denotes the floor function.    I tried this, but I think something went wrong:  Prove by induction. For $n = 1$ \begin{align*} f_1(x) = \frac{1}{(1-1)!} \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} (x-k)^{1-1} &amp;= \sum_{k = 0}^{[x]} (-1)^k \binom{1}{k} =\begin{cases} 1, &amp; 0 \le x &lt; 1 \\ 0, &amp; \text{otherwise} \end{cases} \end{align*} Furthermore \begin{align*} f_{n+1}(x) &amp;= \mathbb P(S_{n+1} = x) = \mathbb P(S_n + X_{n+1} = x) \\ &amp;= \sum_{m = 0}^{\infty} \mathbb P(S_n + X_{n+1} = x \mid X_{n+1} = m) \mathbb P(X_{n+1} = m) \\ &amp;= \sum_{m = 0}^{\infty} \mathbb P(S_n + m = x)  \cdot 1_{[0,1]}(m)  = \mathbb P(S_n = x) + \mathbb P(S_n = x-1) \\ &amp;= \frac{1}{(n-1)!} \sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \frac{1}{(n-1)!} \sum_{k = 0}^{[x-1]} (-1)^k\binom{n}{k}(x-1-k)^{n-1} \\ &amp;=  \frac{1}{(n-1)!}\left(\sum_{k = 0}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 0}^{[x]-1} (-1)^k\binom{n}{k}(x-1-k)^{n-1}\right) \\ &amp;=  \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} + \sum_{k = 1}^{[x]} (-1)^{k-1}\binom{n}{k-1}(x-1-(k-1))^{n-1}\right) \\ &amp;=  \frac{1}{(n-1)!}\left(x^{n-1}+\sum_{k = 1}^{[x]} (-1)^k\binom{n}{k}(x-k)^{n-1} - (-1)^{k}\binom{n}{k-1}(x-k)^{n-1}\right). \end{align*}  How can I solve this?  Edit: \begin{align*} f_{n+1}(x) &amp;= (f_n * f_{X_{n+1}})(x)  = \int_{-\infty}^\infty f_{n}(y) f_{X_{n+1}}(x-y) \, dy \\ &amp;= \int_{-\infty}^\infty \frac{1}{(n-1)!} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(0,1)}(x-y) \, dy \\ &amp;= \frac{1}{(n-1)!} \int_{-\infty}^{\infty} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \cdot 1_{(x-1, x)}(y)\, dy \\ &amp;= \frac{1}{(n-1)!} \int_{x-1}^{x} \sum_{k = 0}^{[y]} (-1)^k \binom{n}{k}(y-k)^{n-1} \, dy \end{align*}  I want to swap the sum and the integral, but the sum depends on $y$.
80966
I define $X_i$ as a random variable that is uniformly distributed between (0,1). What is the expected number of such variables I require to make the sum go just higher than 1.  Thanks
38105
Probably everyone has once come across the following "theorem" with corresponding "proof": $$\sum_{n=0}^\infty 2^n = -1$$ Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.  Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:  If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.  Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.  Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential.
99967
For a positive integer $n$, prove that: $$\sum_{d\mid n} (-1)^{n/d} \cdot \varphi(d) = \begin{cases} 0 &amp; \text{if $n$ is even}  \\ -n  &amp; \text{if $n$ is odd}  \end{cases}$$  I can't even think, how, to begin with, this problem, please provide easy proof of this.
75123
What is the sum of $\sum\limits_{i=1}^{n}ip^i$ and does it matter, for finite n, if $|p|&gt;1$ or $|p|&lt;1$ ?   Edition :  Why can I integrate take sum and then take the derivative ? I think that kind of trick is not always allowed.  ps. I've tried this approach but I made mistake when taking derivative, so I've asked, mayby I should use some program (or on-line app) for symbolic computation.
24958
I found$\ \prod_{i=1}^\infty \frac{p_i}{p_i-1}\approx 11.0453$ on . Moreover, writing a paper, should one provide a proof or it is trivial? Thanks in advance.