= Solution
For $p>1$, the bound $X_k\leq2$ ensures the finite <moment> $m_p=\mathbb EX_1^p\leq2^p$. By <Jensen's inequality>, $m_p\geq(\mathbb EX_1)^p=1$. The process $Z_k=Y_k^p$ is an <adapted>, nonnegative <integrable random variable> at each time, and <independence> gives
$$
\mathbb E[Z_{k+1}\mid\mathcal H_k]
=Y_k^p\mathbb EX_{k+1}^p=m_pZ_k\geq Z_k.
$$
Thus $(Z_k)$ is a <submartingale>, and $\mathbb EZ_n=m_p^n$. The <Doob maximal inequality for a nonnegative submartingale> states that $c\mathbb P(\max_{k\leq n}Z_k\geq c)\leq\mathbb EZ_n$ for $c>0$. Apply it with $c=\lambda^p$. Since the power <function> is increasing on $[0,\infty)$,
$$
\mathbb P\left(\max_{1\leq k\leq n}Y_k\geq\lambda\right)
\leq\frac{m_p^n}{\lambda^p}.
$$
Combining this with the preceding bound gives
$$
\boxed{\mathbb P\left(\max_{1\leq k\leq n}Y_k\geq\lambda\right)
\leq\min\left\{\lambda^{-1},\ m_p^n\lambda^{-p}\right\}.}
$$
The result is a bound on the running maximum, which is why a maximal <inequality> for a <submartingale>, rather than an estimate of the single terminal <random variable>, is needed.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[§4.4, Theorem 4.4.2, printed p. 235: the submartingale maximal inequality]{ref}
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