1

Let $p \in (0,1)$ and $X$ be a random variable such that $P(X=a) = p, P(X=-b) = 1-p$

Show that if $Y$ is another random variable such that $E[X] = E[Y]$ and $V(X) = V(Y)$ then $P(Y \ge a) \le p$ and that equality holds iff $X$ and $Y$ are equal in distribution.

I have a hint to use Chebychev's inequality here but I am not quite getting the desired result.

By Chebychev we have $P(\lvert Y - E[Y] \rvert \ge a ) \le \frac{V(Y)}{a^2}$ which we can re-write with $E[X]$ and $V(X)$ but I am still not getting the result.

I can write it as:

$P(Y \ge a) = P( Y - E[X] \ge a - E[X]) \le P( \lvert Y - E[X] \rvert \ge a - E[X]) \le \frac{V(X)}{(a-E[X])^2} = \frac{p(a+b)^2(1-p)}{(a-E[X])^2} = \frac{p(a+b)^2(1-p)}{(1-p)^2(a+b)^2} = \frac{p}{1-p}$

and $\frac{p}{1-p}$ may be large than $p$ so I haven't achieved the result.

which would be using Chebychev but this doesnt help since $E[X] = ap - b(1-p)$ and $(a-E[X])^2 = [(1-p)(a+b)]^2$ so if say $p = \frac{1}{2}$ then I just get the last part being equal to $1$ which doesn't insure the original is less than or equal to $p$.

oliverjones
  • 61
  • 5
  • 1
    If you recall the proof of Chebyshev's Inequality, this result is contained within it. – whuber Jul 10 '20 at 18:38
  • @whuber I don't immediately see it from the proof of Chebychev's inequality. – oliverjones Jul 10 '20 at 18:43
  • 1
    Read the "comments" section near the bottom of http://www.quantdec.com/envstats/notes/class_06/properties.htm and look at the illustrated proof above it. – whuber Jul 10 '20 at 18:54
  • @whuber I think I see it from that but that still leaves a gap in my understanding how to show this. As you can see above I have applied Chebychev's Inequality but the end result doesn't insure this original quantity is less than $p$ – oliverjones Jul 10 '20 at 19:14
  • 1
    Please do not [cross-post](https://math.stackexchange.com/q/3752435/3212). Decide where to ask. – StubbornAtom Jul 10 '20 at 19:27
  • 1
    For some other approaches, first note that the result does not depend on $a$ and $b$ and consult https://stats.stackexchange.com/questions/45588. – whuber Jul 10 '20 at 19:46
  • @whuber I don't understand why I now need an upper bound on the variance, this doesn't answer my question. I have applied Chebychev's Inequality to the problem and I end with a result that I have shown may be equal to 1 if $p = 1/2$ which does nothing to show the desired result that $P(Y \ge a) \le p $ so how does an upper bound for variance help? – oliverjones Jul 10 '20 at 19:55
  • The point isn't to answer the question, but to show you effective techniques for attacking it. You might be too hung up on using Chebychev's Inequality. – whuber Jul 10 '20 at 20:13
  • @whuber I was given the hint that Chebychev's Inequality would solve it but it doesn't seem to work, nor can I seem to solve it via Markov's Inequality and what you linked gives me an upper bound for variance which makes trying to solve it even more difficult. – oliverjones Jul 10 '20 at 20:20

0 Answers0