st.statistics: minimizes the asymptotic variance of an ergodic average subject to a set of restrictions

Leave

  • $ (E, E math, lambda) $ Y $ (E & # 39 ;, mathcal E & # 39 ;, lambda & # 39;) $ be measuring spaces
  • $ I $ be a finite set not empty
  • $ varphi_i: E & # 39; a E $ be $ ( mathcal E & # 39 ;, mathcal E) $-measurable with $$ lambda & # 39; circ varphi_i ^ {- 1} = q_i lambda tag1 $$ for $ i in I $
  • $ p, q_i: E to (0, infty) $ be $ mathcal E $-measurable with $$ int p : { rm d} lambda = int q_i : { rm d} lambda = 1 $$
  • $ w_i: E a (0,1) $ be $ mathcal E $-measurable, $ w_i & # 39 ;: = w_i circ varphi_i $, $$ p_i & # 39 ;: = begin {cases} frac p {q_i} circ varphi_i & text {on} left {q_i circ varphi_i> 0 right } \ 0 & text {on} left {q_i circ varphi_i = 0 right } end {cases} $$ Y $$ f_i & # 39 ;: = begin {cases} frac f {q_i} circ varphi_i & text {on} left {q_i circ varphi_i> 0 right } \ 0 & text {on} left {q_i circ varphi_i = 0 right } end {cases} $$ for $ i in I $
  • $ zeta $ denote the counting measure in $ (I, 2 ^ I) $ Y $$ nu & # 39 ;: = w & # 39; p & # 39; ( zeta otimes lambda & # 39;) $$

Leave $ f in L ^ 2 ( lambda) $. Assume $$ {q_i = 0 } subseteq {w_ip = 0 }, tag2 $$ $$ {p = 0 } subseteq {f = 0 } tag3 $$ Y $$ {pf ne0 } subseteq left { sum_ {i in I} w_i = 1 right }. tag4 $$

Leave $ ((T_n, X_n & # 39;)) _ {n in mathbb N} $ be the Markov chain (supposed to be in stationarity) generated by the Metropolis-Hastings algorithm with objective distribution $ nu & # 39; $ Y $$ A_n: = frac1n sum_ {i = 0} frac {f & # 39;} {p & # 39;} (T_i, X_i & # 39;) ; ; ; text {for} n in mathbb N. $$ I want to minimize asymptotic variance $$ sigma ^ 2: = lim_ {n to infty} n operatorname {Var} A_n $$ With respect to $ w_i $. How can we do that?

I know that yes $ (Y_n) _ {n in mathbb N_0} $ is any homogeneous Markov chain in time, $ mu: = math L (Y_0) $, $ g in L ^ 2 ( mu) $ Y $ B_n: = frac1n sum_ {i = 0} ^ {n-1} g (Y_i) $, so $ operatorname {Var} B_n = frac1n left ( operatorname {Var} _ mu (g) ​​+2 sum_ {i = 1} ^ {n-1} left (1- frac in right) operatorname {Cov} (f (Y_0), f (Y_i)) right) $. Furthermore, if $ L ^ 2_0 ( mu): = left {h in L ^ 2 ( mu): int h : { rm d} mu = 0 right } $, $ mathcal D (G): = left {h_0 in L ^ 2_0 ( mu): left ( sum_ {i = 0} ^ n kappa ^ ih_0 right) _ {n in mathbb N_0} text {is convergent} right } $, $$ Gh_0: = sum_ {n = 0} ^ infty kappa ^ nh_0 ; ; ; text {for} h_0 in mathcal D (G), $$ Y $ g_0: = g- int g : { rm d} mu in mathcal D (G) $, so $$ n operatorname {Var} B_n xrightarrow {n to infty} 2 langle Gg_0, g_0 rangle_ {L ^ 2 ( mu)} – operatorname {Var} _ mu (g) ​​ tag5. $$ In particular, leaving $ mathcal L: = – (1- kappa) $, we can consider the spectral gap of $ math L $, $$ operatorname {gap} mathcal L = inf _ { substack {h in L ^ 2 ( mu) setminus {0 } \ 1 : perp : h}} frac { langle- mathcal Lh, h rangle_ {L ^ 2 ( mu)}} { left | h right | _ {L ^ 2 ( mu)} ^ 2} = 1- left | kappa right | _ { mathfrak L (L ^ 2_0 ( mu))}, $$ where do we consider $ kappa $ as a non-negative self-attachment operator in $ L ^ 2 ( mu) $. With this definition, the right side of $ (5) $ it is at most $ left ( frac2 { operatorname {gap} mathcal L} -1 right) operatorname {Var} _ mu (g) ​​$.

stochastic processes: variance of a random variable obtained from a linear transformation

Edit: I needed to review this question as suggested.

Suppose there are $ N $ Realizations of the Gaussian process denoted as vectors $ mathbf {z} _ {j} in mathbb {R} ^ {n} $ for $ j = 1, ldots, N $. Leave $ and $ be a random variable such that $ y = sum_ {j = 1} ^ {N} ( mathbf {B} mathbf {z} _ {j}) (i) $
where $ mathbf {B} $ It is a unitary matrix. What is the variance of $ y2?

Explanation: Boldface represents the vector or matrix. $ ( mathbf {B} mathbf {x}) (i) $ represents the $ i $-th vector entry $ mathbf {B} mathbf {x} $.

probability – variance of a fair currency

Consider that Vamshi decides to throw a fair coin repeatedly until he gets a tail. He does almost $ 4 $ draws.The value of the variance $ T $ is ______


I have tried this standard deviation in student marks

$ x —– 1 —– 2 —— 3 —– 4 $

$ P (x) —- frac {1} {2} —– frac {1} {4} —— frac {1} {8} —- frac {1} {16} $

The average is $ frac {1} {4} left ( frac {1} {2} + frac {1} {2 ^ 2} + frac {1} {2 ^ 3} + frac { 1} {2 ^ {} right) = frac {15} {64} $

Then, the variance will be,

$ frac {1} {4} left ( left ( frac {15} {64} – frac {1} {2} right) ^ {2} + left ( frac {15} {64 } – frac {1} {4} right) ^ {2} + left ( frac {15} {64} – frac {1} {8} right) ^ {2} + left ( frac {15} {64} – frac {1} {16} right) ^ {2} right) = $$ frac {460} {16384} $


But the answer is like that.

$ E left (X ^ { right) = 1 ^ {2} times frac {1} {2} + 2 ^ {2} times frac {1} {4} + 3 ^ {2 } times frac {1} {8} + 4 ^ 2 times frac {1} {16} $

$ E left (X right) = 1 times frac {1} {2} +2 times frac {1} {4} +3 times frac {1} {8} +4 times frac {1} {16} $

$ V left (X right) = E left (X ^ {2} right) – left (E left (X right) right) ^ {2} $$ = frac {252} { 256} $


Why is my approach giving incorrect results?

probability – the variance of a sample from a normal population

Please consider the problem and my solution below. I agree with the answer in the back of the book, but somehow, my solution does not seem right to me. I did
Do it the right way?
Move

Issue:
A normal population has a variation of $ 15 $. If samples of size $ 5 $ are extracted from this population, which
you can expect the percentage to have variations (a) less than $ 10 $?
$ 10 $?

Reply:

Leave $ S ^ 2 $ be the variance of the sample and $ n $ be the size of the sample The expression $ nS ^ 2 / sigma ^ 2 $ will have a chi-square distribution with $ 4 $ degrees of freedom.
begin {align *}
sigma ^ 2 & = 15 \
frac {nS ^ 2} { sigma ^ 2} & = frac {5S ^ 2} {15} = frac {S ^ 2} {3} \
S ^ 2 & = 10 \
frac {nS ^ 2} { sigma ^ 2} & = 10/3 \
end {align *}

Using R we find:
pchisq (10/3, df = 4) = 0.496
Therefore the answer is $ 0.496.

probability: variance of the point product of a random binary vector and a constant vector

Leave $ q in mathbb {R} ^ N $ and Z a random binary vector s.t. The sum of its elements is n.
The variance of the statistics. $ Z cdot q = sum ^ N i = 1} Z_i q_i $ It should be $ frac {N-n} {N-1} times n times frac { sum ^ N _ {i = 1} (q_i – bar q) ^ 2} {N} $However, I do not understand how to achieve this equality.
What I do know is that

  1. $ E[Zcdot q] = n bar q $
  2. $ P[Z_i =1, Z_j =1] = frac {n (n-1)} {N (N-1)} $

I also got to this: $ Z cdot q = sum ^ N_ {i = 1} (q_i – bar q) ^ 2 frac {n-1} {N ^ 2} + sum ^ N_ {j = 1} sum ^ N k = 1, k neq j} (q_j – bar q) (q_k – bar q) frac {n} {N} ( frac {n-1} {N-1} – frac {n} {N}) $
but I do not know how to simplify this expression even further. Can someone help me, please?

non-linear optimization: a unit vector that maximizes variance in a discrete probability distribution

This could be a silly question, but I was working on a problem and found the following (sub) problem.

Suppose we have a non-negative vector $ pi in mathbf {R} ^ n $ what satisfies $ sum_ {i = 1} ^ n pi_i = 1 $, that is, it is a discrete probability density. We want to choose a unit vector. $ v in mathbf {R} ^ n $, $ | v | = 1 $, where $ | cdot | $ It is the Euclidean norm, such that "variance".

$ f (v) = sum_ {i = 1} ^ n v_i ^ 2 pi_i – ( sum_ {i = 1} ^ n v_i pi_i) ^ 2 $

it is maximized. Of course there will be multiple maximums, because $ f (v) = f (-v) $.

Are there closed form solutions for the maximum? $ v $ or some idea of ​​how to find it, or what is the maximum value of the function $ f (v) $?
Specifically, I want to show that if $ pi_1> 0.5 $, then any maximum $ v $ satisfy $ { rm sign} (v_1) = – { rm sign} (v_i) $ for all $ i neq 1 $.

Any advice or ideas? Thank you!

performance – Kubernetes – high variance in response times

We have configured the single node of Kubernetes and then the cluster of multiple nodes using kubeadm and we are experiencing performance problems.

We have measured response times and are experiencing variations in response times; Sometimes the execution will be fast and it will be completed in a few seconds, and sometimes the response times will double or triple and, in exceptional cases, they will shoot at a very high irrational value.

  1. I tried to invoke the same endpoint for a period of time. There is almost always an initial period of 10 to 20 seconds when the latency is quite high, after which it normalizes to somewhat better values. However, these APIs are not designed for such a scenario.
  2. A real-world scenario of suede with random reflection times was used and the selected endpoints were invoked for a period (30.60 seconds). In this case there is no pronounced start phase with peaks, but the variation mentioned above is there.

I would appreciate suggestions on what to look for in order to understand and mitigate this problem. When testing the API endpoint, the payload and activity in the cluster remained constant.

We are having the Kubernetes configuration using EC2 in AWS. We have used tools like weavescope and htop and we have not found hunger in terms of CPU or memory.

  • A single node is an EC2 instance of 8 cores and 32 GB.
  • Multi-node is a configuration of three nodes (Master: 4 cores, 16 GB, 8 cores, 32 GB and 4 cores, 16 GB)

probability – Reverse of sum of exponential random variables (mean and variance)

Assume $ X_1, X_2, points, X_n $ following the exponential distribution with mean $ theta> 0 $ and the statistics:
$$
T = sum limits_ {i = 1} ^ n x_i
$$

I know that the sum of the exponential random variables follows the distribution of Gamma, but I can not infer anything about the inverse of the sum $ frac {1} {T} $. My guess would be that:
$$
E left[frac{1}{T}right] = frac {1} { theta} quad text {y} V left[ frac{1}{T} right] = frac {1} { theta ^ 2}
$$

but why?

random variables – Calculate the variance of the population using means and variances of the stratum

I have three strata (1-3) with the mean and variance of 25/5, 26.5 / 4.5 and 24.5 / 10 respectively. I have calculated that the sample sizes are 11, 4 and 25 (40 in total).

How can I calculate the total variance of the population according to the means and variances of the stratum?
My objective is to compare the SE of the SRS and the stratified sample.

Probability: conditional expected value and variance of the average preservation propagation

My first post here, and my math skills are more than a little rusty. I have a simple question for you: suppose that Y is an average spread of X.

Is it always true that: E (X | X> Y) <E (Y | Y <X)? How to prove it?

Is it also true that, for some value of C> 0, Var (Y | Y> C) < Var(X|X>DO)?

Thank you very much in advance!