 Solution

Convergence in distribution makes $(X_n)$ tight. Hence for every $\eta >0$ there is $M$ such that $\mathbb{P}(|X_n|>M)<\eta$ for all sufficiently large $n$. Since $Y_n-c\stackrel{P}{\to }0$,Letting $\eta \downarrow 0$ gives $X_n(Y_n-c)\stackrel{P}{\to }0$. Also, the continuous-mapping theorem gives $c{X}_n\stackrel{d}{\to }cX$. Sincethe Slutsky theorem yields

Now write the requested statistic asThe central limit theorem givesSince $\mathbb{E}[Z_i^2]=1<\infty$, the strong law of large numbers givesalmost surely, and therefore its reciprocal converges in probability to one. Applying the product result provesThis is a self-normalized central limit theorem with a second-moment denominator.