In the probability textbook that I'm reading through right now ("Knowing the Odds, an Introduction to Probability" by John B. Walsh), one of the textbook questions asks us to prove a generalized form of Chebyshev's inequality:
Let $\Phi$ be a strictly positive increasing function, $X$ a random variable, and $\lambda>0$. Prove that: $$\mathbb{P}(\Phi(X)\geq\lambda)\leq\frac{1}{\Phi(\lambda)}\mathbb{E}(\Phi(X))$$
My steps are as follows: we know that $\mathbb{P}(\Phi(X)\geq\lambda) = \mathbb{E}(1_{\Phi(X)\geq\lambda})$. Writing out this latter form as an integral, we have: $$\mathbb{E}(1_{\Phi(X)\geq\lambda}) = \int_{\Phi(X)\geq\lambda}1d\mathbb{P}$$ But at the same time, we can recognize that on the set $\{\Phi(X)\geq\lambda\}$, we have that $\frac{\Phi(X)}{\lambda}\geq 1$. So: $$\int_{\Phi(X)\geq\lambda}1d\mathbb{P}\leq \int_{\Phi(X)\geq\lambda}\frac{\Phi(X)}{\lambda}d\mathbb{P}\leq \int_\Omega\frac{\Phi(X)}{\lambda}d\mathbb{P} = \frac{1}{\lambda}\int_\Omega\Phi(X)d\mathbb{P}$$ What I don't know how to do is turn that $\lambda$ into a $\Phi(\lambda)$. What should my next step be? Any hints would be appreciated. Thanks!
EDIT: it turns out that my textbook has a typo in it. The true generalized Chebyshev's inequality is given by: $$\mathbb{P}(X\geq\lambda) \leq \frac{1}{\Phi(\lambda)}\mathbb{E}(\Phi(X))$$ This follows from a similar line of reasoning as above, and by recognizing that $\frac{\Phi(X)}{\Phi(\lambda)}$, since $\Phi(\cdot)$ is an increasing, positive function.
As an alternative proof, the generalized Chebyshev inequality follows from the basic Chebyshev inequality $$P(|X|\ge\lambda)\le\frac{E|X|^p}{\lambda^p},$$ as follows. Argue that the event $\{X\ge\lambda\}$ is the same as the event $\{\Phi(X)\ge \Phi(\lambda)\}$. Now apply the basic Chebyshev inequality with $p=1$ (aka Markov's inequality) to this last event.