Slutsky's theorem

In probability theory, Slutsky's theorem extends some properties of algebraic operations on convergent sequences of real numbers to sequences of random variables.

The theorem was named after Eugen Slutsky. Slutsky's theorem is also attributed to Harald Cramér.

Statement

Let X n , Y n {\displaystyle X_{n},Y_{n}} be sequences of scalar/vector/matrix random elements. If X n {\displaystyle X_{n}} converges in distribution to a random element X {\displaystyle X} and Y n {\displaystyle Y_{n}} converges in probability to a constant c {\displaystyle c}, then

X n + Y n → d X + c ; {\displaystyle X_{n}+Y_{n}\ {\xrightarrow {d}}\ X+c;}
X n Y n → d X c ; {\displaystyle X_{n}Y_{n}\ \xrightarrow {d} \ Xc;}
X n / Y n → d X / c , {\displaystyle X_{n}/Y_{n}\ {\xrightarrow {d}}\ X/c,} provided that c is non-zero,

where → d {\displaystyle {\xrightarrow {d}}} denotes convergence in distribution.

Notes:

The requirement that Yn converges to a constant is important — if it were to converge to a non-degenerate random variable, the theorem would be no longer valid. For example, let X n ∼ U n i f o r m ( 0 , 1 ) {\displaystyle X_{n}\sim {\rm {Uniform}}(0,1)} and Y n = − X n {\displaystyle Y_{n}=-X_{n}}. The sum X n + Y n = 0 {\displaystyle X_{n}+Y_{n}=0} for all values of n. Moreover, Y n → d U n i f o r m ( − 1 , 0 ) {\displaystyle Y_{n}\,\xrightarrow {d} \,{\rm {Uniform}}(-1,0)}, but X n + Y n {\displaystyle X_{n}+Y_{n}} does not converge in distribution to X + Y {\displaystyle X+Y}, where X ∼ U n i f o r m ( 0 , 1 ) {\displaystyle X\sim {\rm {Uniform}}(0,1)}, Y ∼ U n i f o r m ( − 1 , 0 ) {\displaystyle Y\sim {\rm {Uniform}}(-1,0)}, and X {\displaystyle X} and Y {\displaystyle Y} are independent.
The theorem remains valid if we replace all convergences in distribution with convergences in probability.

Proof

This theorem follows from the fact that if Xn converges in distribution to X and Yn converges in probability to a constant c, then the joint vector (Xn, Yn) converges in distribution to (X, c) (see here).

Next we apply the continuous mapping theorem, recognizing the functions g(x,y) = x + y, g(x,y) = xy, and g(x,y) = x y−1 are continuous (for the last function to be continuous, y has to be invertible).

Slutsky's theorem

Statement

Proof

See also

Further reading