# Finding an integrable function using the MVT

Let $\ (\Omega,F,P)$ a probability space $X, X_n, n=1,2,\ldots$ are real valued random variables on $(\Omega,F,P)$. Assume that $\ E[e^{c|X|}]< \infty$ for some $c>0$. Define $\ X_n = n(e^{X/n}-1), n\geq 1$. By MVT for every $\ n\geq 1$ and every $\omega \in \Omega$ there exists $\ t_n(\omega) \in (0,1/n)$ s.t $\ X_n(\omega)$= $\ X(\omega)$ $\ e^{t_n(\omega)}$. $\ X(\omega)$ choose $\ n_0$ s.t $c>2/n_0$.

Find an integrable random variable $Y$ on $\ (\Omega,F,P)$ s.t $|X_n| \leq Y$ for all $\ n\geq n_0$

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You proved that $|X_n|\leqslant|X|\mathrm e^{|X|/n}$. Note that, for every $x\geqslant0$, $x\leqslant (2/c)\mathrm e^{cx/2}$ hence $|X_n|\leqslant (2/c)\mathrm e^{c|X|/2}\mathrm e^{|X|/n}$. If $n\geqslant 2/c$, $|X|/n\leqslant c|X|/2$ hence $|X_n|\leqslant Y$ with $Y=(2/c)\mathrm e^{c|X|}$ which is integrable.