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X is a Random Variable that can take values only in $[0,10]$. Suppose $P[X>5] \leq \frac{2}{5}$ and $P[X<1]\leq \frac{1}{2}$. Then what is the interval for the $E[X]$?

Answer : $E[X]\geq0.5$ and $E[X]\leq 8.5$.

It is pretty straightforward to show that $E[X]\geq0.5$ via Markov's Inequality but I don't understand how to show the Upper Bound for the interval.

my approach $E[X]\leq 5P[X\leq5] + 10P[X>5] \leq5 * \frac{3}{5} + 10 * \frac{2}{5} = 7$ Could this work?

petearGriffin
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    For a high mean put as much probability as possible at $10$ and the rest at $5$. For a low mean put as much probability as possible at $0$ and the rest at $1$. If you can do the latter, then consider $Y=10-X$ and try to minimise $E[Y]$ in the same way to find an upper bound on $E[10-Y]$ – Henry Mar 29 '22 at 14:57

1 Answers1

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Giving $X$ the largest possible values and probabilities subject to the constraints will maximize its expectation.

Thus, the variable $X$ equal to $10$ with a probability of $2/5$ and equal to $5$ with a probability of $3/5$ will maximize the expectation.

Formally, let $F$ be the distribution for $X,$ given by $F(x)=\Pr(X\le x)$ for $0\le x \le 10.$ The information given you is

  1. For all $\epsilon \gt 0,$ $F(-\epsilon)=0.$ (The probability of the empty set is zero.)

  2. $F(10)=1.$ (The probability of the sample space is $1.$)

  3. For all $\epsilon\gt 0,$ $F(1-\epsilon)\le 1/2 = 0.5.$

  4. $1-F(5) \le 2/5.$ Equivalently, $F(5) \ge 3/5 = 0.6.$

Subject to these constraints, you wish to maximize the expectation, which can be expressed in terms of $F$ as

$$E[X] = \int_0^{10}\left(1 - F(x)\right)\,\mathrm{d}x.$$

The values of $x$ explicitly referenced in the constraints are $0, 1, 5,$ and $10.$ So, let's partition this integral at those points. Here's a picture where the permissible regions for the graph of $F$ are unshaded: that is, the graph must stay within the white area.

Figure 1

Although it is difficult to draw, bear in mind that constraint $(1)$ forces $F=0$ to the left of $x=0$ and constraint $(2)$ forces $F=1$ to the right of $x=10.$

Here is an example of such a distribution function. Bear in mind that it cannot decrease at any point from left to right.

Figure 2

The expectation is the area above the graph of $F$ and to the right of $0.$ To maximize that area, then, we must make $F$ as low as possible subject to the constraints. Here, in red, is the graph of an $F$ that comes close to that optimum. The area represented by the integral for $E[X]$ is shaded in red.

Figure 3

At this point I hope it is obvious that the best possible upper bound for the expectation is the sum of three rectangular areas: from $0$ to $1$ (with height $1$), from $1$ to $5$ (with height $1$), and from $5$ to $10$ (with height $2/5$). Multiplying the widths times the heights, we find

$$E[X] \le (1-0)\times(1) + (5-1)\times(1) + (10-5)\times(2/5) = 7.$$

Incidentally, because $7\le 8.5,$ the statement in the question is correct--but $8.5$ is not the best possible upper bound.

whuber
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  • thanks @whuber i just added my approach could you take a look at it and tell me if that could be true? – petearGriffin Mar 29 '22 at 17:20
  • Yes, it looks valid. Your approach essentially includes a contribution of $2$ from the (white) rectangle between $x=1$ to $x=5$ from the heights of $0$ and $0.5$ in my last figure. Thus, it doesn't account for the fact $F$ must be increasing, but it does account for the fact it mustn't be negative. As such you get an upper bound, but not necessarily the best upper bound. – whuber Mar 29 '22 at 17:24
  • exactly that was my concern because the best possible upper bound could be reduced to a smaller value but the exam wanted us to choose the best possible answer. so it's like the answer $E[X] \leq 8.4$ is not NOT-TRUE but it isn't correct either. – petearGriffin Mar 29 '22 at 17:28
  • Clearly $8.5$ is an upper bound but not the best possible. If the exam insists that $8.5$ is best, then there's something the matter with it. Note that "best possible answer" is not necessarily the same as "best possible" when you are given a limited set of choices. – whuber Mar 29 '22 at 18:14
  • I apologize: upon looking back at the original post, it became clear that your calculation of $7$ is best possible. I have updated the last figure and its explanation to reflect this. The optimum is attained by a variable that has a $3/5$ chance to equal $5$ and a $2/5$ chance to equal $10,$ corresponding to the two jumps in the graph at $x=5$ and $x=10$ as required by constraints $(4)$ and $(2).$ – whuber Mar 29 '22 at 18:45
  • thank you @whuber for helping me out with this question. – petearGriffin Mar 29 '22 at 18:48