1. Continuous Random Variable Proof

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?

2. Re: Continuous Random Variable Proof

Originally Posted by Greenmango
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 $\displaystyle P(X \geq 0)=0$:
$\displaystyle \mu=\int_{0}^{\infty}(x(f(x))dx$
don't you mean

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

3. Re: Continuous Random Variable Proof

Oh sorry, I actually made a typo in the question, it should say P(X <= 0)

4. Re: Continuous Random Variable Proof

Originally Posted by Greenmango
Oh sorry, I actually made a typo in the question, it should say P(X <= 0)
What is 'IT'
Really the whole problem does not make any sense.
Are you given a $PDF$ for the random variable $X~?$
Or you given more about the mean $\mu~?$
This seems to me to be a very strange question.

5. Re: Continuous Random Variable Proof

"It" as in the question at the top marked by "Question:" as well as all the places where I make reference to the question by saying P(X >= 0). I know it's a weird question, that's why I'm having trouble haha, but it's written word for word how my professor gave it to me, no PDF and no mean!

6. Re: Continuous Random Variable Proof

Take a look at the Chebyshev Inequality

7. Re: Continuous Random Variable Proof

Originally Posted by Greenmango
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 $\displaystyle P(X \geq 0)=0$:
$\displaystyle P(X<|\mu t|)=\int^{|\mu t|}_{0}(f(x))dx$
Oh come on. How are we suppose to know what $f(x)$ is?

8. Re: Continuous Random Variable Proof

Originally Posted by Plato
Oh come on. How are we suppose to know what $f(x)$ is?
it's the probability density function of the random variable $X$

I don't think we are supposed to know anything about it other than $X$ has a finite mean and support on the positive real numbers.