I am studying the proof of Markov's inequality in Larry Wasserman's "All of Statistics", shown below:
$$ \mathbb{E}(X)= \int_0^{\infty}xf(x)dx \ge \int_t^{\infty}xf(x)dx \ge t\int_t^{\infty}f(x)dx = tP(X >t)$$
I understand this part:
$$ \mathbb{E}(X)= \int_0^{\infty}xf(x)dx \ge \int_t^{\infty}xf(x)dx$$
I don't understand this:
$$\int_t^{\infty}xf(x)dx \ge t\int_t^{\infty}f(x)dx$$
How does one prove the above step is correct?
Best Answer
You can also use integration by parts. Having $u=x$, and $dv= f(x)dx$, then $uv-\int vdu$ will be equal to $$\left[x\int f(x)dx\right] - \int f(x)dx$$ with the first part evaluate at $t$ and $0$. From here all we need to see is that $\int f(x)dx$ is always positive. So
$$\int xf(x)dx= \left[x\int f(x)dx\right] - \int f(x)dx \ge \left[x\int f(x)dx\right]= t\int f(x)dx.$$