# 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.[1] The theorem was named after Eugen Slutsky.[2] Slutsky's theorem is also attributed to Harald Cramér.[3]

## Statement

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

• $\displaystyle{ X_n + Y_n \ \xrightarrow{d}\ X + c ; }$
• $\displaystyle{ X_nY_n \ \xrightarrow{d}\ Xc ; }$
• $\displaystyle{ X_n/Y_n \ \xrightarrow{d}\ X/c, }$   provided that c is invertible,

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

Notes:

1. 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converges to a constant is important — if it were to converge to a non-degenerate random variable, the theorem would be no longer valid.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 $\displaystyle{ X_n \sim {\rm Uniform}(0,1) }$ and $\displaystyle{ Y_n = -X_n }$. The sum $\displaystyle{ X_n + Y_n = 0 }$ for all values of n. Moreover, $\displaystyle{ Y_n \, \xrightarrow{d} \, {\rm Uniform}(-1,0) }$, but $\displaystyle{ X_n + Y_n }$ does not converge in distribution to $\displaystyle{ X + Y }$, where $\displaystyle{ X \sim {\rm Uniform}(0,1) }$, $\displaystyle{ Y \sim {\rm Uniform}(-1,0) }$, and $\displaystyle{ X }$ and $\displaystyle{ Y }$ are independent.[4]
2. 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 (Xc) (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).