Statistics Graduate Level Quiz: Law Of Large Numbers

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Law Of Large NumbersQuestion 1 of 10

Let $X_1,X_2,\ldots$ be independent with common mean $\mu$ and common finite variance $\sigma^2$ . Consider the weighted average $W_n=\sum_{i=1}^n w_{ni}X_i$ , where $w_{ni}=2i/[n(n+1)]$ . Which statement is correct?

W_n\to(2/3)\mu

in probability because the larger indices receive asymptotically dominant, linearly growing weights.

W_n\to\mu

in mean square because the weights sum to one and

\sum_{i=1}^nw_{ni}^2\sim 4/(3n)\to0

\operatorname{Var}(W_n)\to4\sigma^2/3

because the final few observations each retain weight bounded away from zero.

W_n

need not converge because unequal weights invalidate every weighted law of large numbers for independent observations.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Law Of Large Numbers

Practice Law Of Large Numbers in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Law Of Large Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$W_n\to(2/3)\mu$ in probability because the larger indices receive asymptotically dominant, linearly growing weights.
$W_n\to\mu$ in mean square because the weights sum to one and $\sum_{i=1}^nw_{ni}^2\sim 4/(3n)\to0$ . (correct answer)
$\operatorname{Var}(W_n)\to4\sigma^2/3$ because the final few observations each retain weight bounded away from zero.
$W_n$ need not converge because unequal weights invalidate every weighted law of large numbers for independent observations.

Explanation: When you see a weighted average of i.i.d. random variables, your first instinct should be to check two things: do the weights sum to one, and does

\sum_i w_{ni}^2 \to 0

? These are the key conditions for mean-square (and hence probability) convergence to

\mu

. For

w_{ni} = 2i/[n(n+1)]

, verify the weights sum to one:

\sum_{i=1}^n \frac{2i}{n(n+1)} = \frac{2}{n(n+1)} \cdot \frac{n(n+1)}{2} = 1

. So

E[W_n] = \mu

exactly for every

n

. Now compute the variance:

\operatorname{Var}(W_n) = \sigma^2 \sum_{i=1}^n w_{ni}^2 = \sigma^2 \sum_{i=1}^n \frac{4i^2}{n^2(n+1)^2}

. Using

\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6

, this simplifies to

\sigma^2 \cdot \frac{4(2n+1)}{6n(n+1)} \sim \frac{4\sigma^2}{3n} \to 0

. Since both the bias is zero and the variance vanishes,

W_n \to \mu

in mean square — confirming B. Choice A is wrong because the weights do grow with

i

, but they're normalized so the sum equals one, meaning

E[W_n] = \mu

, not

(2/3)\mu

. The heavier weighting on larger indices affects the rate of variance decay, not the limit. Choice C confuses the behavior of individual weights with the sum of squared weights — no single weight stays bounded away from zero (each

w_{nn} = 2/(n+1) \to 0

), so variance goes to zero, not

4\sigma^2/3

. Choice D is a red herring; unequal weights are perfectly compatible with LLN-type results as long as

\sum w_{ni}^2 \to 0

. Your takeaway: for any weighted average, always check

\sum w_{ni} = 1

(unbiasedness) and

\sum w_{ni}^2 \to 0

(variance collapse). Together, they guarantee mean-square convergence to

\mu

Question 2

Let $X_1,X_2,\ldots$ be iid with a distribution symmetric about zero. For $x\ge e$ , suppose $P(|X_1|>x)=1/(x\log x)$ , with the remaining probability assigned symmetrically on $[-e,e]$ . Which statement about $\overline X_n$ is correct?

$\overline X_n$ converges in distribution to a nondegenerate Cauchy law, because every symmetric distribution with infinite mean belongs to the Cauchy stable domain of attraction.
$\overline X_n\to0$ almost surely because symmetry about zero substitutes for the finite-mean condition in the strong law of large numbers.
$\overline X_n$ cannot converge in probability to any finite constant because a finite absolute first moment is necessary for every weak law of large numbers.
$\overline X_n\to0$ in probability but not almost surely; the truncation conditions for the weak law are satisfied, yet $\sum_nP(|X_n|>\varepsilon n)=\infty$ forces large terms to occur infinitely often. (correct answer)

Explanation: When you encounter a question about sample means with heavy-tailed distributions, your first instinct should be to check whether the mean is finite, then separately assess conditions for the weak law (WLW) versus the strong law (SLLN). Here,

P(|X_1| > x) = 1/(x \log x)

for

x \geq e

. Computing

E|X_1| = \int_e^\infty P(|X_1|>x)\,dx = \int_e^\infty \frac{1}{x\log x}\,dx = \Big[\log\log x\Big]_e^\infty = \infty

, so the mean is infinite. This rules out the classical SLLN immediately. However, the weak law doesn't strictly require finite mean — it requires a truncation condition. Specifically, if

n\,P(|X_1|>n)\to 0

, the WLW holds (with centering). Here

n\,P(|X_1|>n) = n/(n\log n) = 1/\log n \to 0

, so

\overline{X}_n \to 0

in probability (symmetry kills the centering term). Yet for the SLLN, you need

\sum_n P(|X_1|> \varepsilon n) < \infty

, but

\sum_n \frac{1}{n\log n} = \infty

, meaning large values occur infinitely often almost surely — blocking a.s. convergence. This confirms D. Choice A is wrong because not every infinite-mean symmetric distribution attracts to a Cauchy law; the Cauchy domain requires a specific regular variation condition (

P(|X|>x)\sim C/x

), not

1/(x\log x)

. Choice B is wrong because symmetry alone cannot replace finite mean in the SLLN — the Etemadi/Kolmogorov SLLN genuinely requires

E|X_1|<\infty

. Choice C is wrong because it overstates what's needed for the WLW; a finite mean is sufficient but not necessary. Study tip: Always distinguish the WLW truncation condition (

n\,P(|X|>n)\to 0

) from the SLLN's summability requirement (

\sum_n P(|X|>\varepsilon n)<\infty

) — they can diverge, as this problem perfectly illustrates.

Question 3

Let $\{X_i\}_{i\ge 1}$ be a weakly stationary sequence with mean $\mu$ and autocovariance function $\gamma(h)$ . Suppose $\sum_{h=0}^{\infty}|\gamma(h)|<\infty$ . Which conclusion about $\overline X_n=n^{-1}\sum_{i=1}^n X_i$ follows directly from these assumptions?

$\overline X_n\to\mu$ in mean square and hence in probability, but almost-sure convergence is not implied by the covariance condition alone. (correct answer)
$\overline X_n\to\mu$ almost surely because absolute summability of the autocovariances is sufficient for the strong law.
$\sqrt n(\overline X_n-\mu)\to0$ in probability because the variance of the normalized sum is of order $n^{-2}$ .
$\overline X_n$ need not converge in probability because weak stationarity permits nonzero correlation between distant observations.

Explanation: When you see a question combining weak stationarity with an absolute summability condition on autocovariances, your first instinct should be to compute

\text{Var}(\overline{X}_n)

directly — this reveals exactly what mode of convergence is guaranteed. Under weak stationarity,

\text{Var}(\overline{X}_n) = n^{-2}\sum_{i,j}\gamma(i-j) = n^{-1}\sum_{h=-(n-1)}^{n-1}\left(1-\frac{|h|}{n}\right)\gamma(h)

. Since

\sum_{h=0}^{\infty}|\gamma(h)|<\infty

, the Cesàro-weighted sum is bounded, so

\text{Var}(\overline{X}_n) = O(n^{-1}) \to 0

. This means

\overline{X}_n \to \mu

in mean square, and since mean-square convergence implies convergence in probability, both follow. This confirms A is correct. B is wrong because absolute summability of autocovariances is a second-moment (covariance) condition. Almost-sure convergence requires stronger tools — typically independence, mixing rates, or moment conditions beyond what

\gamma(h)

alone provides. You cannot promote an

L^2

result to an a.s. result for free. C contains a subtle arithmetic error.

\text{Var}(\overline{X}_n) = O(n^{-1})

, not

O(n^{-2})

, so

\text{Var}(\sqrt{n}(\overline{X}_n - \mu)) = O(1)

— it stays bounded, not vanishing. Hence

\sqrt{n}(\overline{X}_n - \mu)

does not go to zero in probability. D is wrong because absolute summability of

\gamma(h)

is precisely the condition that controls long-range correlations enough to ensure convergence. The correlations decay sufficiently fast. Study tip: Always anchor stationarity-and-averaging questions to the variance calculation of

\overline{X}_n

. The mode of convergence you can claim is determined entirely by what controls that variance —

L^2

conditions give

L^2

/probability convergence, not almost-sure convergence.

Question 4

An analyst observes iid pairs $\{(X_i,Y_i)\}_{i=1}^n$ with finite second moments and $\operatorname{Var}(X_i)>0$ . The analyst fits an ordinary least-squares regression of $Y_i$ on an intercept and $X_i$ . No linear conditional-mean model is assumed.

To what quantity does the fitted slope $\widehat\beta_n$ converge almost surely?

$E[\partial E(Y_i\mid X_i)/\partial X_i]$ , the average derivative of the conditional mean function.
$E(Y_i)/E(X_i)$ , the ratio of the population means whenever the denominator is nonzero.
$E(Y_i/X_i)$ , the average observation-level slope whenever the ratio is almost surely defined.
$\operatorname{Cov}(X_i,Y_i)/\operatorname{Var}(X_i)$ , the coefficient of the population linear projection of $Y_i$ on $X_i$ . (correct answer)

Explanation: When you see OLS applied without assuming a correctly specified linear model, the key question becomes: what does OLS actually estimate in large samples? This question tests your understanding of the population linear projection, a concept distinct from conditional expectation. By the Law of Large Numbers, the OLS slope satisfies

\widehat\beta_n = \frac{\frac{1}{n}\sum(X_i - \bar{X})(Y_i - \bar{Y})}{\frac{1}{n}\sum(X_i - \bar{X})^2} \xrightarrow{a.s.} \frac{\operatorname{Cov}(X_i, Y_i)}{\operatorname{Var}(X_i)}

. This limit is the slope of the best linear predictor of

Y_i

given

X_i

— the population linear projection — and it holds regardless of whether

E[Y_i \mid X_i]

is actually linear. Answer D is therefore correct. Choice A is tempting but wrong. The average derivative

E[\partial E(Y_i \mid X_i)/\partial X_i]

equals the OLS limit only when

E[Y_i \mid X_i]

is linear in

X_i

, which is not assumed here. In general, these two quantities differ. Choice B,

E(Y_i)/E(X_i)

, confuses the ratio of means with a regression coefficient — this would be relevant for ratio estimation, not OLS. Choice C,

E(Y_i/X_i)

, has no algebraic connection to the OLS formula and reflects a naive "pointwise slope" idea that doesn't arise from minimizing squared prediction error. The key study takeaway: always anchor OLS asymptotics to the linear projection formula

\operatorname{Cov}(X,Y)/\operatorname{Var}(X)

. This is what OLS recovers in the population, model misspecification or not.

Question 5

For each $n$ , let $X_{n1},\ldots,X_{nn}$ be independent within row, with $X_{ni}=n$ with probability $1/n$ and $X_{ni}=0$ otherwise. Define $\overline X_n=n^{-1}\sum_{i=1}^nX_{ni}$ . Although $E(\overline X_n)=1$ for every $n$ , what is the limiting behavior of $\overline X_n$ ?

$\overline X_n\to1$ in probability, because every row consists of identically distributed variables with finite expectations.
$\overline X_n\to0$ in probability, because each individual variable is zero with probability approaching one.
$\overline X_n\Rightarrow\operatorname{Poisson}(1)$ , so it does not converge in probability to its expectation. (correct answer)
$\overline X_n\Rightarrow N(1,1)$ , so asymptotic normality replaces convergence in probability.

Explanation: Whenever you see a triangular array where individual entries grow with

n

, you should immediately think about Poisson convergence, not classical LLN behavior. The key diagnostic: as

n \to \infty

, each

X_{ni}

is nonzero with probability

1/n \to 0

, but there are

n

such terms — a classic rare-event accumulation setup. To see why C is correct, count how many of the

n

variables actually equal

n

. Let

S_n = \#\{i : X_{ni} = n\}

. Each indicator is Bernoulli

(1/n)

, so

S_n \sim \text{Binomial}(n, 1/n)

, which converges in distribution to

\text{Poisson}(1)

. Now notice that

\overline{X}_n = n^{-1}\sum X_{ni} = n^{-1} \cdot n \cdot S_n = S_n

. So

\overline{X}_n \stackrel{d}{=} S_n \Rightarrow \text{Poisson}(1)

. The mean of

\text{Poisson}(1)

is 1, consistent with

E(\overline{X}_n)=1

, but the distribution is not degenerate at 1 — so convergence in probability to the mean fails entirely. Choice A is the core trap: it invokes the LLN, but the LLN's i.i.d. version requires a fixed distribution. Here the distribution of

X_{ni}

changes with

n

, so classical LLN does not apply. Choice B confuses individual behavior with aggregate behavior — yes, each term is zero with high probability, but the aggregate

\overline{X}_n

still fluctuates because rare large values dominate. Choice D misapplies the CLT; with variance

\text{Var}(\overline{X}_n) = 1

not shrinking to zero, there is no normal concentration around the mean. Your study tip: whenever individual probabilities vanish like

1/n

but the number of trials grows like

n

, Poisson limits govern the sum — not Gaussian CLT or LLN concentration.

Question 6

Let $X_1,X_2,\ldots$ be iid with $E|X_1|<\infty$ and mean $\mu$ . For each $n$ , let $N_n$ be a positive integer-valued random variable independent of the entire sequence, with $N_n\to\infty$ in probability. Define $T_n=N_n^{-1}\sum_{i=1}^{N_n}X_i$ . What can be concluded under the stated assumptions?

$T_n$ need not converge in probability, because the random sample size does not satisfy $N_n/n\to1$ .
$T_n\to\mu$ almost surely, because the ordinary strong law remains valid under every random subsequence.
$T_n\to\mu$ in probability, while almost-sure convergence need not follow from convergence of $N_n$ in probability. (correct answer)
$T_n\to0$ in probability, because division by the diverging random sample size dominates the partial sum.

Explanation: When a sample size is itself random, you need the random index law of large numbers: if

N_n \to \infty

in probability and

N_n

is independent of the

X_i

's, then

T_n = N_n^{-1}\sum_{i=1}^{N_n} X_i \to \mu

in probability. The key tool is a standard epsilon-delta argument conditioning on

N_n

. For any

\varepsilon > 0

, the event

\{|T_n - \mu| > \varepsilon\}

has small probability whenever

N_n

is large (by the ordinary LLN applied at each fixed sample size), and since

N_n \to \infty

in probability, you can make that probability arbitrarily small. This confirms C is correct. Almost-sure convergence, however, is a strictly stronger mode of convergence and requires more structure. Knowing only that

N_n \to \infty

in probability is insufficient to upgrade from "in probability" to "almost surely" — you would need

N_n \to \infty

a.s. (or a summability condition on the tails). So C correctly identifies the precise strength of the conclusion. A is wrong because convergence in probability of

T_n

does not require

N_n/n \to 1

; it only requires

N_n \to \infty

in probability, which is given. B overclaims: almost-sure convergence of

T_n

does not follow automatically from the ordinary SLLN when the index is only assumed to diverge in probability. D is wrong because the partial sum

\sum_{i=1}^{N_n} X_i

grows like

\mu N_n

, so dividing by

N_n

recovers

\mu

, not zero. The study tip: always distinguish carefully between convergence modes. In probability and almost surely are not interchangeable — know what each assumption about

N_n

buys you.

Question 7

Let $X_1,X_2,\ldots$ be pairwise independent and identically distributed, and suppose $E|X_1|<\infty$ . No mutual independence or finite variance is assumed. Which conclusion is valid?

Only $\overline X_n\to E(X_1)$ in probability is guaranteed; almost-sure convergence requires mutual independence.
$\overline X_n\to E(X_1)$ almost surely; pairwise independence and integrability suffice for this strong law. (correct answer)
$\overline X_n\to E(X_1)$ almost surely only if the common distribution additionally has finite variance.
No law of large numbers follows, because pairwise independence does not control any variance or tail behavior.

Explanation: When a question asks what the Law of Large Numbers guarantees under pairwise independence, your instinct might be to reach for the classical strong law, which typically assumes mutual independence. But a deeper result exists that many students overlook. The key theorem here is Etemadi's Strong Law of Large Numbers (1981): if

X_1, X_2, \ldots

are pairwise independent and identically distributed with

E|X_1| < \infty

, then

\overline{X}_n \to E(X_1)

almost surely. Etemadi's proof cleverly avoids relying on mutual independence or finite variance — it uses truncation arguments and the pairwise independence structure alone to establish a.s. convergence. This makes B correct: pairwise independence plus integrability is genuinely sufficient for the strong law. A is wrong because it understates what's possible. It correctly notes convergence in probability but falsely claims a.s. convergence requires mutual independence — Etemadi's theorem directly contradicts this. C is wrong for a similar reason: finite variance is a sufficient condition for the classical strong law via Kolmogorov's criterion, but it is not necessary. Assuming

E|X_1^2| < \infty

is a much stronger requirement than

E|X_1| < \infty

, and the theorem doesn't need it. D is wrong because it conflates "no variance control" with "no conclusion." Variance-based arguments (like Chebyshev) are just one route; Etemadi's approach sidesteps them entirely. Study tip: Watch for questions that test whether you know the minimal conditions for major theorems. Examiners often use distractors that cite a sufficient condition (like finite variance or mutual independence) and present it as necessary — a classic trap on graduate-level probability exams.

Question 8

Suppose $X_1,X_2,\ldots$ are independent and identically distributed with $P(X_i>x)=x^{-3/2}$ for $x\ge1$ . Which statement about the sample mean is correct?

$\overline X_n\to3$ almost surely; the mean is finite (equal to 3) even though $\operatorname{Var}(X_i)=\infty$ , and the strong law requires only $E|X_i|<\infty$ . (correct answer)
$\overline X_n\to3$ only in probability, because infinite variance prevents the almost-sure version of the law of large numbers.
$\overline X_n$ diverges almost surely, because a Pareto distribution with tail exponent $3/2<2$ has no finite population mean.
$\overline X_n\to1$ almost surely, because the lower endpoint of the support determines the long-run average when the variance is infinite.

Explanation: When you see a question involving a heavy-tailed distribution and the sample mean, your first instinct should be to carefully compute whether

E[X_i]

is finite — because the Strong Law of Large Numbers (SLLN) requires only

E|X_i| < \infty

, nothing more. Here,

P(X_i > x) = x^{-3/2}

for

x \geq 1

defines a Pareto-type distribution with density

f(x) = \frac{3}{2}x^{-5/2}

. Computing the mean directly:

E[X_i] = \int_1^\infty x \cdot \frac{3}{2}x^{-5/2}\,dx = \frac{3}{2}\int_1^\infty x^{-3/2}\,dx = \frac{3}{2}\cdot\left[\frac{x^{-1/2}}{-1/2}\right]_1^\infty = \frac{3}{2}\cdot 2 = 3.

Since

E[X_i] = 3 < \infty

, the SLLN guarantees

\overline{X}_n \to 3

almost surely. The variance being infinite is irrelevant — the SLLN only cares about the first moment. This confirms A is correct. B is wrong because it conflates the SLLN with the Central Limit Theorem. Infinite variance disables normal CLT approximations, but it does not downgrade the SLLN from almost-sure to in-probability convergence. C is wrong because it misapplies the "tail exponent" rule. The mean fails to exist only when the tail exponent satisfies

\alpha \leq 1

(i.e.,

P(X>x) \sim x^{-\alpha}

with

\alpha \leq 1

). Here

\alpha = 3/2 > 1

, so the mean is finite. D is a fabrication — the support's lower endpoint has no special role in determining the long-run average. Study tip: Always check finiteness of

E|X_i|

before worrying about variance. The SLLN is robust — infinite variance is a CLT problem, not an LLN problem.

Question 9

Let $I_1,I_2,\ldots$ be independent Bernoulli random variables satisfying $P(I_n=1)=1/n$ . Which statement best describes the convergence of $I_n$ ?

$I_n\to0$ in probability but not almost surely, because successes occur infinitely often with probability one. (correct answer)
$I_n\to0$ almost surely, because the individual success probabilities converge to zero.
$I_n$ converges neither in probability nor almost surely, because the series of success probabilities diverges.
$I_n\to1$ almost surely, because independence and the divergent probability sum force eventual success.

Explanation: When studying convergence of random variables, you need to keep two distinct modes of convergence carefully separated: convergence in probability and convergence almost surely. The key tools here are the Borel-Cantelli Lemmas. Since

P(I_n = 1) = 1/n

, notice that

P(I_n = 1) \to 0

, which immediately implies

I_n \to 0

in probability — for any

\varepsilon > 0

P(|I_n| > \varepsilon) = 1/n \to 0

. So far so good. Now ask whether convergence is almost sure. Almost sure convergence of

I_n \to 0

would require that, with probability one,

I_n = 1

happens only finitely often. The Second Borel-Cantelli Lemma says: if events are independent and

\sum P(I_n = 1) = \sum 1/n = \infty

, then

P(I_n = 1 \text{ i.o.}) = 1

. Since

I_n = 1

occurs infinitely often with probability one,

I_n

cannot converge to 0 almost surely. This confirms A is correct. Now for the distractors. B is wrong because

p_n \to 0

is sufficient for convergence in probability but not for almost sure convergence — you need the series of probabilities to also converge (First Borel-Cantelli), which fails here. C is wrong because divergence of

\sum p_n

blocks almost sure convergence but does not block convergence in probability, which holds regardless. D is nonsensical —

I_n \to 1

would require

p_n \to 1

, the opposite of what's given. Your study tip: always check both Borel-Cantelli lemmas when independence is mentioned. Divergence of

\sum p_n

plus independence is the signature of "in probability but not almost surely."

Question 10

Suppose $X_i$ are independent Bernoulli random variables with $P(X_i=1)=i^{-1/2}$ , and let $\overline X_n=n^{-1}\sum_{i=1}^nX_i$ . Which limiting statement is correct?

$\overline X_n\to0$ almost surely, because the average success probability vanishes and the centered averages satisfy a strong law. (correct answer)
$\overline X_n\to1/2$ almost surely, because the success probabilities have the same order as a harmonic sequence.
$\overline X_n$ fails to converge almost surely, because the expected total number of successes is infinite.
$\overline X_n\to0$ only in probability, because the variables are independent but not identically distributed.

Explanation: When dealing with the strong law for non-identically distributed independent random variables, your first instinct should be to check Kolmogorov's strong law: if

\sum_{i=1}^n \text{Var}(X_i)/i^2 < \infty

, then

\overline{X}_n - \overline{\mu}_n \to 0

almost surely, where

\overline{\mu}_n = n^{-1}\sum_{i=1}^n \mu_i

. Here,

\mu_i = E[X_i] = i^{-1/2}

, so

\overline{\mu}_n = n^{-1}\sum_{i=1}^n i^{-1/2}

. Since

\sum_{i=1}^n i^{-1/2} \sim 2\sqrt{n}

, we get

\overline{\mu}_n \sim 2n^{-1}\sqrt{n} = 2n^{-1/2} \to 0

. For the variance condition:

\text{Var}(X_i) \leq 1

, so

\sum_{i=1}^\infty \text{Var}(X_i)/i^2 \leq \sum_{i=1}^\infty i^{-2} < \infty

. Kolmogorov's criterion is satisfied, confirming

\overline{X}_n \to 0

almost surely. Answer A is correct. Answer B is wrong because it conflates

i^{-1/2}

with harmonic-type behavior. The harmonic series

\sum i^{-1}

diverges at rate

\log n

, but

\sum i^{-1/2} \sim 2\sqrt{n}

, so

\overline{\mu}_n \to 0

, not

1/2

. Answer C confuses the divergence of

\sum \mu_i = \sum i^{-1/2} = \infty

with failure of the average to converge — but it's the average that matters, and averages can still converge even when totals diverge. Answer D incorrectly claims that non-identical distributions block almost sure convergence; the SLLN generalizes well beyond the i.i.d. setting via Kolmogorov's criterion. Your study tip: whenever variables are independent but not identically distributed, immediately check Kolmogorov's variance condition

\sum \text{Var}(X_i)/i^2 < \infty

and separately compute

\overline{\mu}_n

. These two pieces together fully determine almost sure behavior of

\overline{X}_n