Abstract: Let the randomized query complexity of a relation for error probability ϵ be denoted by Rϵ(⋅). We prove that for any relation f⊆0,1<sup>n</sup>×R and Boolean function g:0,1<sup>m</sup>→0,1, R1/3(f∘g<sup>n)</sup>=Ω(R4/9(f)⋅R1/2−1/n<sup>4(g)), where f∘g<sup>n is the relation obtained by composing f and g. We also show that R1/3(f∘(g<sup>⊕O(log</sup>n))<sup>n)=Ω(log</sup>n⋅R4/9(f)⋅R1/3(g)), where g<sup>⊕O(log</sup>n) is the function obtained by composing the xor function on O(logn) bits and g<sup>t.