Question:If X is a random variable for which $\displaystyle P(X \geq 0)=0$ and $\displaystyle |\mu_x|<\infty$, show that $\displaystyle P(X<\mu t) \geq 1 - \frac{1}{t}$ for every $\displaystyle t \geq 1$.

My TA told me to use the following information:

$\displaystyle \mu=\int_{-\infty}^{\infty}(x(f(x))dx$

Since $\displaystyle P(X \geq 0)=0$:

$\displaystyle \mu=\int_{0}^{\infty}(x(f(x))dx$

$\displaystyle \mu=\int_{0}^{|\mu t|}(x(f(x))dx + \int_{|\mu t|}^{\infty}(x(f(x))dx$

My own work:

We know that:

$\displaystyle P(X<\mu t)=\int^{\mu t}_{-\infty}(f(x))dx$

Since $\displaystyle P(X \geq 0)=0$:

$\displaystyle P(X<|\mu t|)=\int^{|\mu t|}_{0}(f(x))dx$

We also know:

$\displaystyle \begin{align*}1&=\int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx \\

1 - \frac{1}{t}&=\int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t}\end{align*}$

So we have to show:

$\displaystyle \begin{align*}

P(X<\mu t) &\geq 1 - \frac{1}{t} \\

\int_{0}^{|\mu t|}(f(x))dx &\geq \int_{0}^{|\mu t|}(f(x))dx + \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t} \\

0 &\geq \int_{|\mu t|}^{\infty}(f(x))dx- \frac{1}{t} \\

\frac{1}{t} &\geq \int_{|\mu t|}^{\infty}(f(x))dx

\end{align*}$

This looks similar to what we had in the formula for $\displaystyle \mu$ except for the extra x in the integrand, so I assume this is where that comes in, but I'm not sure how to relate the two.

Does anyone know where I should go from here? Or am I taking a wrong turn?