My Math Forum help with probability question

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 October 24th, 2017, 02:29 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 97 Thanks: 2 help with probability question The question: Could anyone give me a tip on how to approach this question? Thank you. Last edited by skipjack; October 24th, 2017 at 03:56 AM.
 October 24th, 2017, 01:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,586 Thanks: 612 What is X and what is its distribution?
October 24th, 2017, 01:08 PM   #3
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Quote:
 Originally Posted by mathman What is X and what is its distribution?
I got the impression it doesn't matter as long as X is a legit random variable.

I'm showing that given the assumption made on $f(x)$ that

$E[f(x)] = f(E[x])$

but I've probably made an error somewhere.

 October 25th, 2017, 04:26 PM #4 Global Moderator   Joined: May 2007 Posts: 6,586 Thanks: 612 Let $\displaystyle f(x)=x^2$ Let X be a random variable with 2 values 0 and .5, each with probability .5. Then E(X)=.25 while E(f(X))=.125. You need to say something about the distribution of X.
 October 25th, 2017, 06:04 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 $\mu,~\eta \text{ are real constants.}$ $X$ is the only random object here. $f(X) = f(\mu)+ (X-\mu)f^\prime(\mu)+\dfrac 1 2 (X-\mu)^2 f^{\prime\prime}(\eta)$ $E[f(X)] = f(\mu) + f^\prime(\mu)E[X] - \mu f^\prime(\mu) + \dfrac 1 2 f^{\prime \prime}(\eta) E[X^2] - \mu f^{\prime \prime}(\eta) E[X] + \dfrac {\mu^2 f^{\prime \prime}(\eta)}{2}$ $E[f(X)] =E[X^2] \dfrac 1 2 f^{\prime \prime}(\eta) + E[X]\left( f^\prime(\mu) - \mu f^{\prime \prime}(\eta)\right)+\left( f(\mu) - \mu f^\prime(\mu) + \dfrac {\mu^2 f^{\prime \prime}(\eta)}{2} \right)$ now $f(E[X]) = f(\mu) + (E[X]-\mu)f^\prime(\mu) + \dfrac 1 2 (E[X]-\mu)^2 f^{\prime\prime}(\eta)$ ah.. I see where my error arose. $f(E[X]) =\dfrac{1}{2} (E[X])^2 f''(\eta )+E[X] \left(f'(\mu )-\mu f''(\eta )\right) +\left(f(\mu )-\mu f'(\mu )+\dfrac{1}{2} \mu ^2 f''(\eta )\right)$ $E[f(X)] - f(E[X]) =\dfrac 1 2 f^{\prime \prime}(\eta) \left(E[X^2] - (E[X])^2\right)= \dfrac 1 2 f^{\prime \prime}(\eta) \sigma_X^2$ as $f^{\prime \prime}(\eta) \geq 0 \text{ and }\sigma_X^2 \geq 0$ $E[f(X)] - f(E[X]) \geq 0$ $E[f(X)] \geq f(E[X])$

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