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Studying for quals... This is from Durrett, problem 2.13. If $Y_n \geq 0$, $EY_n^\alpha \to 1$ and $E Y_n^\beta \to 1$, for some $0 <\alpha < \beta$, then $Y_n \to 1$ in probabiltiy.

I note that by Jensen + squeezing $EY_n^\gamma \to 1$ for all $\gamma \in [\alpha,\beta]$.

Intuitively speaking $EY_n^\alpha \to 1$ says that the mass of $Y_n$ is either near $1$ or somehow $\alpha$-power symmetric. Since $\alpha$-power symmetry and $\beta$-power symmetry are different, the hypotheses intuitively imply that $Y_n$ must have its mass near $1$.

Can someone send me in the right direction?

Edit: The hypotheses imply that $\{ Y_n^\alpha \}$ is bounded in $L^{\beta/\alpha}$ and hence a uniformly integrable family since $\beta/\alpha>1$.

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Hint: For $t > 0$, $\beta t^\alpha - \alpha t^\beta \le \beta - \alpha$, with equality only at $t=1$.

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    $\begingroup$ Okay I've got it. Given $\epsilon > 0$ choose $\delta$ so that $\beta t^\alpha - \alpha t^\beta \leq \beta - \alpha - \delta$ whenever $|t-1| > \epsilon$. Then $E[\beta Y_n^\alpha + \alpha Y_n^\beta] \leq (\beta-\alpha-\delta)P(|Y-1|>\epsilon) + (\beta-\alpha)P(|Y-1| \leq \epsilon)$. The LHS tends to $\beta - \alpha$, so the weight $P(|Y-1| > \epsilon)$ must tend to $0$. Now my REAL question. How in the name of great intuition do you decide to consider $\beta t^\alpha - \alpha t^\beta$? $\endgroup$
    – nullUser
    Aug 6, 2015 at 2:39
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    $\begingroup$ Given $\beta t^\alpha - \alpha t^\beta \leq \beta - \alpha$ it was straightforeward to finish the problem. But why consider that function, and why consider bounding it by $\beta - \alpha$? Those are two leaps that I don't think I would have made. $\endgroup$
    – nullUser
    Aug 6, 2015 at 2:43
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    $\begingroup$ You know about $\mathbb E [Y^\alpha]$ and $\mathbb E[Y^\beta]$, and therefore of any linear combination. What linear combination of $t^\alpha$ and $t^\beta$ would be maximized or minimized at $t=1$? Take the derivative... $\endgroup$ Aug 6, 2015 at 5:16
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The inequality suggested is wrong, it only works for $t<1$. So you can't argue with that bound with delta

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