If I know the probabilities of a non-negative random variable constraints $P(X < 3) = \frac{1}{3}$ and $P(X \geq 6) = \frac{1}{6}$, how then I find all possible expected values E[X]? I tried to use Markov Inequality for the second probability to then find that $E[X] \geq 1$, $X \geq 6$. I struggle to find its values if $X < 6$.
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Hint: show that the possible $\mathbb{E}X$ are $[m,+\infty]$ for an $m\in\mathbb{R}$ that you should calculate. – user10354138 Feb 09 '21 at 11:06
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Consider the three classes of value with the corresponding probabilities
$[0;3)$ with probability $1/3$
$[3;6)$ with probability $1/2$
$[6;+\infty)$ with probability $1/6$
looking at the classes, we can easy substitute $+\infty=9$, taking the center value of each class, and calcualate the expectation
$$\mathbb{E}[X]=1.5\cdot\frac{1}{3}+4.5\cdot\frac{1}{2}+7.5\cdot\frac{1}{6}=4$$
the possible values of the expectation are in the second class:
$$\mathbb{E}[X] \in [3;6)$$
Of course, in a very theoretically scenario, if you get that the third class is $[6; >33)$ the expectation can be in the third class... but considering that $83.3\%$ of the distribution is less than 6, the more consistent answer is that the expectation is in the second class of values...
Always theoretically speaking you could have an extreme situation with Expectation =2.5 when the first class is totally concentered in 0, the second in 3 and the third in 6, so a general answer is
$$\mathbb{E}[X]\geq 2.5$$
tommik
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But why does the probability $\frac{1}{2}$ for the range of values [3,6). Thank you. – chroniclesofanoctupus Feb 09 '21 at 11:36
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@chroniclesofanoctupus: simply considering that $P(\Omega)=1$ thus $P(3\leq X<6)=1-1/3-1/6=1/2$ – tommik Feb 09 '21 at 11:37
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