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October 24th, 2017, 03:29 AM   #1
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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.
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October 24th, 2017, 02:00 PM   #2
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What is X and what is its distribution?
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October 24th, 2017, 02:08 PM   #3
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Quote:
Originally Posted by mathman View Post
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.
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October 25th, 2017, 05:26 PM   #4
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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.
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October 25th, 2017, 07:04 PM   #5
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$\mu,~\eta \text{ are real constants.}$

$X$ 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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