Expected value derivation question

notgoodatmath · #1 $Old$ December 27th, 2008, 08:52 PM

Hi,

T is a function, $T = \frac{\sum X_n}{\sqrt{n}}$

where, X are i.i.d Bernoulli r.v

The end point of the derivation is,

$Prob(T \ge z) \le e^{-az} \left(\frac{e^{\frac{a}{\sqrt{n}}}+e^{\frac{-a}{\sqrt{n}}}}{2}\right)^n$

I am given,

$Prob(T \ge z) \le e^{-az} (Expectation[e^{aT}])$ , so

$Prob(T \ge z) \le e^{-az} (Expectation[e^{\frac{a.X_i}{\sqrt{n}}}])$

From a previous step, I know that Prob(X = 1) = 0.5, and Prob(X = -1) = 0.5

So the Expected values are

$e^{\frac{a.0.5}{\sqrt{n}}}$ for X = 1, and $e^{\frac{a.-0.5}{\sqrt{n}}}$ for X=-1

So the $Expectation[e^{aT}]$ =
$\frac{e^{\frac{a}{\sqrt{n}}} + e^{\frac{-a}{\sqrt{n}}}}{2}$

At this point it looks like the answer should be

$e^{-az}\left(\frac{e^{\frac{a}{\sqrt{n}}} + e^{\frac{-a}{\sqrt{n}}}}{2}\right)$

I can't see where the nth power around the brackets should come from.

ie. $e^{-az} \left(\frac{e^{\frac{a}{\sqrt{n}}}+e^{\frac{-a}{\sqrt{n}}}}{2}\right)^n$

Any ideas?

Last_Singularity · #2 $Old$ December 27th, 2008, 10:31 PM

Quote:

Originally Posted by notgoodatmath $View Post$

Hi,
I am given,

$Prob(T \ge z) \le e^{-az} (Expectation[e^{aT}])$ , so

$Prob(T \ge z) \le e^{-az} (Expectation[e^{\frac{a.X_i}{\sqrt{n}}}])$

You're close! Except you missed one thing - observe the quoted text: when you plugged in your expression for what T equals, you inputted $\frac{X_i}{\sqrt{n}}$ instead of $\frac{\sum X_n}{\sqrt{n}}$ . That would give you the result for one of such random variable, not but n independently identically distributed of such random variables! Thus, accounting for the one missing component, we should instead have:

$Prob(T \ge z) \le e^{-az} (E[e^{aT}])$
$= Prob(T \ge z) \le e^{-az} (E[e^{a\frac{\sum X_n}{\sqrt{n}}}])$
$= Prob(T \ge z) \le e^{-az} (E[e^{a\frac{X_1 + X_2 + ... + X_n}{\sqrt{n}}}])$
$= Prob(T \ge z) \le e^{-az} (E[e^{a\frac{X_1}{\sqrt{n}}} e^{a\frac{X_2}{\sqrt{n}}} ... e^{a\frac{X_n}{\sqrt{n}}}])$
$= Prob(T \ge z) \le e^{-az} (E[(e^{a\frac{X_1}{\sqrt{n}}})^n])$

And there is your power of n!

notgoodatmath · #3 $Old$ December 27th, 2008, 10:54 PM

thanks for finding my n!