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Slutsky's theorem

by , PhD

Slutsky's theorem concerns the convergence in distribution of the transformation of two sequences of random vectors, one converging in distribution and the other converging in probability to a constant.

Table of Contents

Joint convergence in distribution

Slutsky's theorem is based on the fact that if a sequence of random vectors converges in distribution and another sequence converges in probability to a constant, then they are jointly convergent in distribution.

Proposition (Joint convergence) Let [eq1] and [eq2] be two sequences of random vectors. If [eq3] and [eq4], where $c$ is a constant, then[eq5]

Proof

In the proposition above we have indicated convergence in probability by [eq6] and convergence in distribution by [eq7].

The theorem

We provide a statement of Slutsky's theorem that is slightly more general than the statement usually found in standard references.

Proposition (Slutsky) Let [eq8] and [eq2] be two sequences of random vectors such that [eq3] and [eq11], where $c$ is a constant. Let [eq12] be a continuous function. Then,[eq13]

Proof

The couple [eq14] is jointly convergent in distribution to $left( X,c
ight) $ by the proposition above (Joint convergence). Therefore, by the Continuous Mapping theorem, the fact that [eq15] is continuous implies that [eq16] converges in distribution to [eq17].

The theorem is valid also when [eq18] and [eq19] are sequences of random matrices (the reason being that random matrices can be thought of as random vectors whose entries have been re-arranged into several columns).

Implications

Since the sum and the product are continuous functions of their operands, Slutsky's theorem implies that[eq20]when [eq3], [eq22] and the dimensions of X_n and $Y_{n}$ are such that their sum and/or their product are well-defined.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let [eq18] be a sequence of Kx1 random vectors such that[eq24]where X is a normal random vector with mean mu and invertible covariance matrix V.

Let [eq25] be a sequence of $L	imes K$ random matrices such that[eq26]where A is a constant matrix. Find the limit in distribution of the sequence of products [eq27].

Solution

By Slutsky's theorem[eq28]where[eq29]The random vector Y has a multivariate normal distribution, because it is a linear transformation of a multivariate normal random vector (see the lecture entitled Linear combinations of normal random variables). The expected value of Y is[eq30]and its covariance matrix is[eq31]Therefore, the sequence of products [eq32] converges in distribution to a multivariate normal random vector with mean $Amu $ and covariance matrix $AVA^{intercal }$.

Exercise 2

Let [eq18] be a sequence of Kx1 random vectors such that[eq24]where X is a normal random vector with mean 0 and invertible covariance matrix V.

Let [eq35] be a sequence of $K	imes K$ random matrices such that[eq36]Find the limit in distribution of the sequence[eq37]

Solution

By the Continuous Mapping theorem[eq38]Therefore, by Slutsky's theorem[eq39]Using the Continuous mapping theorem again, we get[eq40]Since V is an invertible covariance matrix, there exists an invertible matrix Sigma such that[eq41]Therefore,[eq42]where we have defined[eq43]The random vector Z has a multivariate normal distribution, because it is a linear transformation of a multivariate normal random vector (see the lecture entitled Linear combinations of normal random variables). The expected value of Z is[eq44]and its covariance matrix is[eq45]Thus, Z has a standard multivariate normal distribution (mean 0 and variance I) and[eq46]is a quadratic form in a standard normal random vector. So, [eq47] has a Chi-square distribution with [eq48] degrees of freedom. In summary, the sequence [eq49] converges in distribution to a Chi-square distribution with K degrees of freedom.

Exercise 3

Let everything be as in the previous exercise, except for the fact that now X has mean mu. Find the limit in distribution of the sequence[eq50]where [eq51] is a sequence of Kx1 random vectors converging in probability to mu.

Solution

Define[eq52]By Slutsky's theorem[eq53]where[eq54]is a multivariate normal random variable with mean 0 and variance V. Thus, we can use the results of the previous exercise on the sequence[eq55]which is the same as[eq56]and we find that it converges in distribution to a Chi-square distribution with K degrees of freedom.

References

van der Vaart, A. W. (2000) Asymptotic Statistics, Cambridge University Press.

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