Convergence of Random Variables

[Glass]

Hi,

What is meant by "convergence of random variables"? Specifically, this statement confuses me:

The sequence of random variables X_1, X_2, ... , X_n is said to converge in probability to the constant c if for any \epsilon > 0:
\lim_{n \rightarrow \infty} P(\vert X_n - c \vert < \epsilon) = 1

 

[matt grime]

it means exactly what it says: for any e>0 the sequence of numbers

P(|X_n - c|<e)

converges to 1. What is confusing about it?

So, if X_n was the normal distribution with mean c and standard deviation 1/n, that would satisfy the condition.

Just work through a few examples, try to figure out what is going on.

As to why this is important, I can't help you, but there ought to be no mystery as to what the definition is: the r.v.s are getting 'closer' to being the constant r.v. X with P(X=c)=1.

 

[Glass]

Sorry I realize now I didn't make myself very clear. What I don't get is, what is this sequence of random variables? And how is it approaching something? For example, consider the case of rolling say a 3 with a fair die. What is X_1? X_2? in this case? I would think there is only one random variable here, so what is this sequence of random variables? And what and how is it approaching?

 

[matt grime]

X_1 and X_2 are nothing (a priori) to do with rolling a 3 with a fair die.

They're just random variables. That's all. They don't represent any real life experiment.

X_1 is an r.v. X_2, is an r.v. etc.

If they satisfy this condition they are said to converge in probability to that constant r.v.

It's just a definition. I gave you an example: X_n is a normal r.v. with mean c and variance 1/n.

If the word sequence bothers you, just think of a family of r.v.s indexed by the integers.

Given such a family, and an e>0, and a c, I can write down a seqeunce of numbers

x_n=P(|X_n-c|<e)

if this sequence of numbers x_n tends to 1, we are in the situation above. X_n, as n increases gets closer to being 'like a constant r.v.' in this case.

If you want examples with die, then X_n as the score on a die, ain't going to converge to anything. If I set X_n to be the average score on n dice then it will converge in probability to the constant r.v. with c=3.5, by the strong law of large numbers, or the central limit theorem, or whatever.

 

[EnumaElish]

Sorry I realize now I didn't make myself very clear. What I don't get is, what is this sequence of random variables? And how is it approaching something? For example, consider the case of rolling say a 3 with a fair die. What is X_1? X_2? in this case? I would think there is only one random variable here, so what is this sequence of random variables? And what and how is it approaching?

Rightly confused you are: a random variable is neither. It is not a variable, it is a (probability) function. Usually a probability function is a deterministic function with known parameters and a known "shape." A sequence of random variables is a sequence of probability functions F₁, ... F_n; P(|X_n - c| < ε) = F_n(ε+c) if x > c; P(|X_n - c| < ε) = 1 - F_n(ε-c) if x < c (where x is the "generic" argument of F_n evaluated "near c").

"A random variable converging to a constant" is, I think, a probability distribution converging to a degenerate distribution "in probability" (pointwise) although not necessarily "in distribution" (uniformly).

 

[Jason Swanson]

Rightly confused you are: a random variable is neither. It is not a variable, it is a (probability) function. Usually a probability function is a deterministic function with known parameters and a known "shape." A sequence of random variables is a sequence of probability functions F₁, ... F_n; P(|X_n - c| < ε) = F_n(ε+c) if x > c; P(|X_n - c| < ε) = 1 - F_n(ε-c) if x < c (where x is the "generic" argument of F_n evaluated "near c").

"A random variable converging to a constant" is, I think, a probability distribution converging to a degenerate distribution "in probability" (pointwise) although not necessarily "in distribution" (uniformly).

This is not right. A real-valued random variable is a measurable function from the probability space to the reals. If the random variable is X, then its distribution (a.k.a. cumulative distribution function) is F(x) = P(X \le x). The objects X and F are different.

Saying X_n \to X in probability means that

P(|X_n - X| &gt; \varepsilon) \to 0

as n\to\infty for each fixed \varepsilon&gt;0. Saying X_n\to X in distribution means that

F_n(x)\to F(x)

as n\to\infty for all x such that F is continuous at x.

In general, convergence in probability implies convergence in distribution, but not conversely. However, if X is a constant (as in the case being discussed), then the two concepts are equivalent.

 

[EnumaElish]

Thanks for your input, Jason.

Isn't X_n ---> X in probability the same thing as F_n(x) ---> F(x) pointwise? If this is right, then I don't think in terms of convergence there is a meaningful difference between X_n ---> X and F_n(x) ---> F(x). For all ends and purposes they are one and the same.

 

[Jason Swanson]

Isn't X_n ---> X in probability the same thing as F_n(x) ---> F(x) pointwise?

These are not the same thing. If F_n(x) \to F(x) pointwise, then F_n(x) \to F(x) for all x such that F is continuous at x. The latter statement is true iff X_n \to X in distribution. Convergence in distribution does not imply convergence in probability.

Here are some examples. Let X_n = 1/n and X = 0. Then X_n \to X in probability and in distribution. But F_n(0) = P(X_n \le 0) = 0 for all n, whereas F(0) = P(X \le 0) = 1. So F_n does not converge to F pointwise.

Let X_1,X_2,\ldots be iid with X_1\sim N(0,1). Let S_n=X_1+\cdots+X_n and Y_n=S_n/\sqrt{n}. Let Y\sim N(0,1). Then Y_n\to Y in distribution. (In fact, Y_n=Y in distribution for all n.) But, for m&lt;n,

Y_n - Y_m = \frac{S_n - S_m}{\sqrt{n}} - \left({\frac1{\sqrt{m}}<br /> - \frac1{\sqrt{n}}}\right)S_m.

If my algebra is correct, this means Y_n - Y_m is normal with mean zero and variance 2(1-\sqrt{m/n}). It follows that \{Y_n\} is not Cauchy in probability, so it cannot converge in probability.

 

[EnumaElish]

I was going with "Convergence in probability is, indeed, the (pointwise) convergence of probabilities."

Are you saying that "(pointwise) convergence of probability" is different from pointwise convergence of F_n(x) in the real analysis sense? I guess that's what you are saying.

Are you also saying that convergence in distribution is different from F_n ---> F in real analysis?

If convergence in dist. is the real analytic F_n ---> F, and assuming "(pointwise) convergence of probability" is the pointwise convergence of F_n(x) to F(x), then the former should have implied the latter. Since it doesn't (except when converging to a constant, see below) then either or both of these are misconceptions on my part. Thanks for pointing this out.

Also, "If Xn converges in distribution to a constant c, then Xn converges in probability to c." Am I right to state "at least when converging to a constant, convergence in distribution implies convergence in probability"?

I do not understand

It follows that {Y_n} is not Cauchy in probability, so it cannot converge in probability.

How is Cauchy part of this? Isn't Cauchy the ratio of two normals?

 

[Jason Swanson]

I was going with "Convergence in probability is, indeed, the (pointwise) convergence of probabilities."

Are you saying that "(pointwise) convergence of probability" is different from pointwise convergence of F_n(x) in the real analysis sense? I guess that's what you are saying.

I do not know what the author(s) of that Wikipedia article mean by "(pointwise) convergence of probability". But I am saying that convergence in probability is different from pointwise convergence of F_n(x).

Are you also saying that convergence in distribution is different from F_n ---> F in real analysis?

Convergence in distribution is equivalent to F_n(x) -> F(x) for all x such that F is continuous at x. This is, in fact, weaker than pointwise convergence.

Also, "If Xn converges in distribution to a constant c, then Xn converges in probability to c." Am I right to state "at least when converging to a constant, convergence in distribution implies convergence in probability"?

Yes, this is correct.

How is Cauchy part of this? Isn't Cauchy the ratio of two normals?

I meant "Cauchy" in the sense of converging sequences, not in the sense of the Cauchy distribution. For the real analysis analogue, a sequence {a_n} is convergent iff it is Cauchy, i.e. for all e > 0, there exists N > 0 such that |a_n - a_m| < e whenever n > N and m > M.

The probabilistic version is this: Let \{Y_n\} be a sequence of random variables. There exists a random variable Y such that Y_n\to Y in probability if and only if \{Y_n\} is Cauchy in probability, i.e. for all \varepsilon&gt;0 and \delta&gt;0, there exists N&gt;0 such that

P(|Y_n - Y_m| &gt; \varepsilon) &lt; \delta

whenever n,m&gt;N.

 

[EnumaElish]

In light of the comments by Jason Swanson, I am going to update my first post under this thread:

A random variable is not a variable, it is a function from the probability space (W) to the set of real numbers (R). And its randomness is completely described by a deterministic probability function with known parameters.

I. Convergence of probability functions (real analysis):
The probability function F: R ---> [0,1] associated with a random variable can be analyzed using the ordinary tools of real analysis. Thus, it can be stated that a sequence of probability functions converge to a limiting probability function, either pointwise (weaker) or uniformly (stronger).

I.A. Pointwise convergence (real analysis):
The sequence {F_n(x)} pointwise converges to F(x) iff for every ε > 0, there is a natural number N_x such that all n ≥ N_x, |F_n(x) − F(x)| < ε.

I.B. Uniform convergence (real analysis):
The sequence {F_n} uniformly converges to F iff for every ε > 0, there is a natural number N such that for all x and all n ≥ N, |F_n(x) − F(x)| < ε.

I.C. Implication (real analysis):
Uniform convergence implies pointwise convergence.

II. Convergence of random variables (probability analysis):
The convergence of random variables is a related but different concept.

II.A. Convergence in distribution (probability analysis):
The weakest concept of convergence of a r.v. is convergence in distribution. A sequence of r.v.'s {X_n} described by respective prob. functions {F_n} is said to converge in distribution to a r.v. X described by a prob. function F iff {F_n(x)} ---> F(x) (pointwise, in the real analytic sense) for all x at which F is continuous.

II.B. Convergence in probability (probability analysis):
Convergence in probability is stronger. {X_n} converges in probability to X iff P(|X_n - X| > ε) ---> 0 for every ε > 0. This is the concept of convergence used in the weak law of large numbers.

II.C. Almost sure convergence (probability analysis):
Still stronger is almost sure convergence. {X_n} converges almost surely to X iff P(X_n ---> X) = 1.

II.D. Sure convergence or "convergence everywhere" (probability analysis):
The strongest concept of probabilistic convergence is sure convergence (convergence everywhere). {X_n} converges surely (or converges everywhere) to X iff {X_n(w)} ---> X(w) (pointwise, in the real analytic sense) for all w in W.

II.E. Implications (probability analysis):
1. Sure convergence implies almost sure convergence.
2. Almost sure convergence implies convergence in probability.
3. Convergence in probability implies convergence in distribution.

II.F. Concepts yet to be invented:
1. Uniform sure convergence: {X_n} converges "uniformly surely" to X iff {X_n} ---> X uniformly in the real analytic sense.
2. Uniform convergence in distribution: A sequence of r.v.'s {X_n} described by respective prob. functions {F_n} is said to "uniformly converge in distribution" to a r.v. X described by a prob. function F iff {F_n} ---> F uniformly in the real analytic sense.

 

[samirrk]

Here is an example of sequence of random variables (which was the original question)

Consider a telephone switchboard that can handle up to 100 simultaneous calls. Define a random variable X as the number of active calls at a given instant of time. Now this random variable has a probability mass function that describes the probability of the following:
0 active calls,
1 active call,
2 active calls and so on up to
100 active calls.

Now consider a related random variable X1 which is the number of active calls at 10:00:00 am. And another random variable X2 which is the number of active calls at 10:00:01 and so on till X900 which is the number of active calls at 10:15:00.

X1 to X900 is a sequence of random variables. They all have the same probability mass function. And we could make the simplifying assumption that they are all independent. In that case, this sequence of random variables is independent, identically distributed (iid).