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Markov's inequality states:

Let $X$ be a non-negative random variable and suppose that $\mathbb{E}(X)$ exists. For any $t > 0$:

$$\mathbb{P}(X > t) \leq \frac{\mathbb{E}(X)}{t}$$

My text contains the following proof:

Since $X > 0$,

$$ \begin{align} \mathbb{E}(X) &= \int_0^{\infty} xf_X(x)dx \\ &= \int_0^{t} xf_X(x)dx + \int_t^{\infty} xf_X(x)dx \\ &\geq \int_t^{\infty} xf_X(x)dx \\ &\geq t\int_t^{\infty} f_X(x)dx \\ &= t\mathbb{P}(X > t) \end{align} $$

My concern is the step where we remove $x$ from the integral. I think the assumption is that since we know $x$ is non-negative, that removing multiplication by $x$ can only make things smaller. However, for $0 < x < 1$ removing multiplication by $x$ should actually make them bigger. If $f_X$ only has density in that range, then I don't think you can say that $\int_t^{\infty} xf_X(x)dx \geq \int_t^{\infty} f_X(x)dx$. Unless somehow multiplying by $t$ at the same time gets rid of this problem?

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Note that the range of integration is $x = t$ to $x=\infty$. On that range, you have $x \geq t$, so you can lower bound $x$ by $t$. Since the density is always nonnegative, $f_X(x) \geq 0$ for any $x$, you can write $xf_X(x) \geq tf_X(x)$ whenever $x \geq t$.

To elaborate, the calculation is $$ \int_t^\infty xf_X(x) dx \geq \int_t^\infty tf_X(x)dx = t \int_t^\infty f_X(x)dx. $$

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  • $\begingroup$ Thanks. I had accepted this proof a while back and came back to it and wasn't sure if I hadn't noticed the "problem" before or if I was missing a detail now. Makes sense. $\endgroup$ Apr 18 '20 at 1:33

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