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The recurrence relation at $k=3$ gives $I_5=2\theta I_3=2{\theta }^2$. ConsequentlyBy variance additivity for independent random variables, the variance of their sample mean is

Substitute the linear regression model into the estimators and use the centring and orthogonality conditions:A linear combination of independent normal random variables is normal. Their means are ${\beta }_0$ and ${\beta }_1$, while variance additivity for independent random variables givesConsequently

The centered design gives $\hat{\beta }=\beta +S_{xx}^{-1}{\sum }_i{x}_i{ϵ}_i$. By independence of the errors, their off-diagonal covariances vanish. Thus variance additivity for independent random variables givesA larger spread of predictor values decreases the slope variance for fixed error variance.

The meeting position $N$ is the sum of $F$ independent Bernoulli random variables, each of mean $p$ and variance $pq$. Linearity of expectation and variance additivity for independent random variables giveThese are the mean and variance of the unwrapped integer label specified in the question, rather than circular angular statistics.

Use the deterministic weights $a_{n,k}=\sqrt{k}/n$. Their squared sum and maximum areTo apply the quadratic expansion of a cumulant-generating function rigorously, write $\log M_X(u)=u^2/2+r(u)$, with $r(u)=o(u^2)$. Fix $t\ne 0$ and letAll arguments $t\sqrt{k}/n$ lie in this shrinking interval. ConsequentlyTaking logs of the moment-generating function and using independence of random variables now givesThe case $t=0$ is immediate. Since $e^{t^2/4}$ is the moment-generating function of a normal distribution of variance $1/2$, the convergence theorem for moment-generating functions givesThis is a weighted central limit theorem with negligible maximum weight. Its limit variance also agrees with the exact value $\mathrm{Var}(T_n)=(n+1)/(2n)$ from variance additivity for independent random variables.

For each random variable, the probability mass function givesLet $S_m={\sum }_{i=1}^m{X}_i$. By linearity of expectation and variance additivity for independent random variables,The independence is needed for the variance calculation, but not for the mean.

Let $x_n$ be a centered scalar beam-deviation coordinate in a deliberately controlled toy model. Let the readout be $s_n=x_n+{\eta }_n$, where the centered measurement noise ${\eta }_n$ has variance $N$ and is independent of $x_n$ and of earlier noise. Stipulate a correctly timed correction of strength $g$: $x_{n+1}=x_n-g{s}_n$. This scalar observation is a teaching surrogate, not a claim that real cooling measures each particle. From variance additivity for independent random variables,For fixed current $V>0$ and $N\ge 0$, differentiate the next-step variance with respect to feedback gain:The unique one-step minimizing feedback gain and resulting variance areThe noisier the readout relative to the current spread, the smaller the useful correction. The exact one-step cooling condition follows by subtraction:Thus excessive positive feedback gain heats this model. The threshold depends on the current variance: a gain that initially cools can later heat if the beam has become narrower or the noise has grown.

For chosen dimensionless $V_0=1$ and $N=1/4$, $g_{\ast }=0.8$ gives $V_1=0.20$. Taking $g=1.8$ instead gives $V_1=1.45$, larger than the initial spread. Holding $g=0.8$ fixed givesso the variance approaches the nonzero noise floor $1/6$. More generally, for $0<g<2$ the fixed-gain limit is $V_{\infty }=gN/(2-g)$. This long-run stability condition differs from the condition for cooling the current beam on the very next step.

If the toy feedback gain is recomputed after every step and $N>0$, the reciprocal recurrence is even simpler:With the chosen inputs, $V_5=1/21\simeq 0.047619$. This ideal model has no additional diffusion, finite observation time or lost correlation; its indefinitely falling spread is not a prediction for an operating storage ring.

To expose the missing physics mathematically, replace $s=x+\eta$ by an arbitrary centered sample signal $s$. Defining $C=\mathrm{Cov}(x,s)$ and $S=\mathrm{Var}(s)>0$ givesUnwanted contributions from other particles or electronics can enlarge $S$ without increasing useful $C$; poor timing or lost correlation can reduce $C$ or even reverse its sign. This is why real stochastic cooling needs measured beam fluctuations, appropriate beam mixing and correctly phased feedback control, rather than an imagined catalogue of individual particle positions.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$ and expected value $\mu =\mathbb{E}[X_1]$. The weak law of large numbers states that their sample mean has convergence in probability to $\mu$: for every fixed $\epsilon >0$,Thus a large sample is unlikely to give an average with a specified fixed-size error. The strong law of large numbers makes the stronger statement of almost sure convergence along an entire sequence. Under the displayed assumptions both laws apply; their names distinguish the conclusions, not two competing predictions.

When the variance ${\sigma }^2$ is finite, variance additivity for independent random variables gives $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. The Chebyshev inequality immediately supplies a useful quantitative form and a brief proof:For independent trials with a Bernoulli distribution of parameter $p=0.2$, the expected value is $0.2$ and the variance is $0.16$. At $n=1000$, the chance of an error of at least $0.1$ is therefore at most $0.016$. The exact binomial distribution gives a much smaller probability, as the figure under the strong law of large numbers shows. The bound is conservative, but it is valid for every sample size under these assumptions and does not require a normal approximation.

The weak law of large numbers addresses a fixed tolerance $\epsilon$. The central limit theorem instead resolves errors on the shrinking scale $\sigma /\sqrt{n}$ and gives their approximate normal distribution. In particular, the weak law of large numbers alone supplies neither a bell-shaped error law nor a universal rate of approach. For example, a Pareto distribution with minimum one and shape $3/2$ has expected value three but infinite variance: both the weak law of large numbers and strong law of large numbers apply, while the finite-variance statement of the central limit theorem does not.