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X is a random variable that can take values only in $[0,10]$. Suppose $P[X>5]\le2/5$ and $P[X<1]\le0.5$. Using Markov's inequality, I found the lower bound of $E[X]$ as $E[X]\ge1\cdot P[X\ge1]\ge0.5$. But how can I find the upper bound of $E[X]$?

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    $\begingroup$ Consider the random variable $X$ that has $P(X=10)=0.4$ and $P(X=5)=0.6$. $\endgroup$ – angryavian Jul 21 at 21:05
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Consider the following random variable $Y = 1 * \mathbf{1}_{\left\{X < 1\right\}} + 5 * \mathbf{1}_{\left\{1 \leq X \leq 5\right\}} + 10 * \mathbf{1}_{\left\{X > 5\right\}}$. By construction $ X \leq Y$ almost surely thus $E[X] \leq E[Y]$. Computing $E[Y]$ we get $$ E[Y] = 1 * P[X < 1] + 10* P[X > 5] + 5 *P[1\leq X \leq 5] \leq 0.5 + 4 + 0.5 = 5, $$ which gives us a bound on $E[X]$.

To see that this bound is sharp given your assumptions, for $ \varepsilon > 0$ consider the random variable $X_\varepsilon$ satisfying $P[X_\varepsilon = 1 - \varepsilon] = 0.5$, $P[X_\varepsilon = 10] = 0.4$, and $P[X_\varepsilon = 5] = 0.1$. $X_\varepsilon$ satisfies all the given assumptions and $E[X_\varepsilon] = 5 - \varepsilon/2$.

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