# Upper and lower bound of E[X] [closed]

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 $$E[X]$$?

New contributor
Anubhab Haldar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Consider the random variable $X$ that has $P(X=10)=0.4$ and $P(X=5)=0.6$. – angryavian Jul 21 at 21:05

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 $$E[X_\varepsilon] = 5 - \varepsilon/2$$.