Q) Suppose $Y_n\geq 0$, $EY_{n}^{\alpha}\rightarrow 1$ and $EY_n^{\beta}\rightarrow 1$ for some $0<\alpha<\beta$. Show that $Y_n\rightarrow 1$ in probability. It has been answered here and here but wanted to check if I can apply and understand tightness.
If there is a $\phi$, s.t. $\phi(x)\rightarrow \infty$ as $|x|\rightarrow \infty$ and $\text{sup}_n\int \phi(x)dF_n(x)<\infty$, then $F_n$ is tight. Does $EY_n^{\alpha}\rightarrow 1\implies$ that the sequence is tight? Firstly, I don't know if $Y_n^{\alpha}(w)\rightarrow \infty$ as $|w|\rightarrow \infty$ and even if it does, not sure how that implies $EY_n^{\alpha}\leq C<\infty$ and what stops $EY_n^{\alpha}$ from blowing up for small $n$?
But if the sequence is tight, rest falls in place. Let $F_{n_k}\implies F$ and since the sequence is tight, $F$ is the distribution function of some random variable Y. There is a theorem that says if $g,h$ are continuous, $|h(x)|/g(x)\rightarrow 0 \text{ as } |x|\rightarrow \infty, F_n\implies F \text{ and } \int g(x)dF_n(x)\leq C< \infty$, then $\int h(x)dF_n(x)\rightarrow \int h(x)dF(x)$. Thus $EY_{n_k}^{\alpha}\rightarrow EY^{\alpha}=1$. By squeezing, $EY^{\gamma} = 1 = (EY^{\alpha})^{\gamma/\alpha}$ for $\gamma \in (\alpha,\beta)$. Thus for the function $f(x) = x^{\gamma/\alpha}$, we have an equality in Jensen's and thus $Y^{\alpha}=EY^{\alpha}=1$. $F_{n_k}\implies F$ and $Y=1=\text{ constant}$ implies $Y_n$ converges to $1$ in probability.
Best Answer
$Y_n \geq 0$ and $EY_n^{\alpha} \to 1$ for some $\alpha >0$ implies that $\{Y_n\}$ is tight: there is a constant $C$ such that $EY_n^{\alpha} \leq C$ for all $n$. By Chebychev's inequality we get $P(Y_n >M)=P(Y_n^{\alpha} >M^{\alpha}) \leq \frac {EY_n^{\alpha}} {M^{\alpha}} \leq \frac C {M^{\alpha}}$. Given $\epsilon >0$ we can find $M$ such that $\frac C {M^{\alpha}} <\epsilon$ and this makes $P(Y_n >M)<\epsilon$ for all $n$.