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This is an exercise in text R. Durrett, Probability: Theory and Examples, in the section entitled "Weak convergence".

Suppose $Y_n\geq0,EY_n^\alpha\rightarrow 1$ and $EY_n^\beta\rightarrow 1$ for some $0<\alpha<\beta$. Show that $Y_n\rightarrow 1$ in probability.

It seems reasonable that $Y_n$ is close to 1 for the following reason. We have $EY_n^\beta\geq(EY_n^\alpha)^{\beta/\alpha}$, because $x\to x^{\beta/\alpha}$ is convex. Here we actually have the "equality" when $n$ is big.

Maybe we should also link it to "weak convergence" in the proof since it's in that section... Please look at it for me. Thank you.

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Homework? See the FAQ. – cardinal May 29 '12 at 22:31
We have $EY_n^\beta\geq(EY_n^\alpha)^{\beta/\alpha}$, because $x\to x^{\beta/\alpha}$ is convex. Here we actually have the equality when $n$ is big, so it seems reasonable that $Y_n$ be close to 1. – Iew May 29 '12 at 23:04
up vote 3 down vote accepted

Fix $\beta\gt\alpha\gt0$ and introduce $\gamma=\beta/\alpha\gt1$. For every $\varepsilon\gt0$, the strict convexity of the function $x\mapsto x^{\gamma}$ and the continuity of the function $x\mapsto x^{\alpha}$ at $x=1$ imply that there exists $c_{\varepsilon}\gt0$ such that, for every $x$, $$ x^\beta=(x^\alpha)^\gamma\geqslant1+\gamma\cdot(x^\alpha-1)+c_{\varepsilon}\cdot[|x-1|\geqslant\varepsilon]. $$ One applies this to $x=Y_n/E(Y_n^\alpha)^{1/\alpha}$ and takes the expectations. This cancels the $\gamma\cdot(x^\alpha-1)$ term, hence $$ E(Y_n^\beta)\geqslant E(Y_n^\alpha)^{\gamma}(1+c_{\varepsilon}\mathrm P(A_n)), $$ with $$ A_n=[|Y_n-E(Y_n^\alpha)^{1/\alpha}|\geqslant\varepsilon E(Y_n^{\alpha})^{1/\alpha}]. $$ Since $E(Y_n^\beta)\to1$, $E(Y_n^\alpha)^{\gamma}\to1$ and $c_{\varepsilon}\gt0$, $\lim\limits_{n\to\infty}\mathrm P(A_n)=0$. Since $E(Y_n^{\alpha})^{1/\alpha}\to1$, for every $n$ large enough, $A_n\supseteq [|Y_n-1|\geqslant2\varepsilon]$. Finally, for every $\varepsilon\gt0$, $\lim\limits_{n\to\infty}\mathrm P(|Y_n-1|\geqslant2\varepsilon)=0$. QED.

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