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October 24th, 2017, 03:29 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 84 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 04:56 AM. 
October 24th, 2017, 02:00 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,435 Thanks: 562 
What is X and what is its distribution?

October 24th, 2017, 02:08 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,755 Thanks: 899  
October 25th, 2017, 05:26 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,435 Thanks: 562 
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, 07:04 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,755 Thanks: 899 
$\mu,~\eta \text{ are real constants.}$ $X$ 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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