# Chebyshev's inequality

• July 15th 2009, 07:56 AM
roshanhero
Chebyshev's inequality
If X is a non-negative random variable for which E(X) exists,show that for every t>0.
$P[X\geq t] \leq\frac {E(X)}{t}$
• July 15th 2009, 10:45 AM
CaptainBlack
Quote:

Originally Posted by roshanhero
If X is a non-negative random variable for which E(X) exists,show that for every t>0.
$P[X\geq t] \leq\frac {E(X)}{t}$

$\frac{E(X)}{t}=\int_0^{\infty} \frac{x}{t}p(x)\; dx = \int_0^t\frac{x}{t}p(x)\; dx + \int_t^{\infty}\frac{x}{t}p(x)\; dx$

......... $\ge \int_t^{\infty}\frac{x}{t}p(x)\; dx$

and as $\frac{x}{t}\ge 1$ for $x\ge t$ hence:

$\frac{E(X)}{t}\ge \int_t^{\infty}\frac{x}{t}p(x)\; dx\ge \int_t^{\infty}p(x)\; dx$

CB
• July 15th 2009, 09:27 PM
roshanhero
How can we prove chebyshev inequality from this relation?
• July 15th 2009, 09:28 PM
matheagle
And you didn't notice that under Mr Fantasy's link...
http://people.csail.mit.edu/ronitt/COURSE/S07/lec25.pdf
under your last post...
http://www.mathhelpforum.com/math-he...nequality.html
it's Theorem ONE, on page ONE.
• July 15th 2009, 09:38 PM
CaptainBlack
Quote:

Originally Posted by roshanhero
How can we prove chebyshev inequality from this relation?

Put $X=(Y-\mu)^2$

CB
• July 15th 2009, 09:45 PM
roshanhero
Thanks,but,what should i use in t.
• July 15th 2009, 10:22 PM
CaptainBlack
Quote:

Originally Posted by roshanhero
Thanks,but,what should i use in t.

try it, see what you get then refer th Chebyshev's inequality and see what you have to do with t to get the required form.

CB