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I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $\mathbb E(X)=20$. How do I find the upper bound to $P(X\le -10)$?

Jimmy R.
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puffles
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2 Answers2

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Hint:

  • $50-X$ is a nonnegative random variable since $50$ is an upperbound.

  • Express your inequality in the form of $Pr(50-X \ge c)$.

Siong Thye Goh
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  • So it should be like this : P(50-X >= -10) ? – puffles Nov 14 '18 at 11:23
  • $P(X \le -10) = P(-X \ge 10) = P(50-X \ge 60)$, now apply Markov on $50-X$. – Siong Thye Goh Nov 14 '18 at 11:24
  • Makes sense. Thanks a bunch! – puffles Nov 14 '18 at 11:25
  • I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability. – puffles Nov 15 '18 at 15:26
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    yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same. – Siong Thye Goh Nov 15 '18 at 15:29
  • helloworld's answer is correct, the mean of the new variable is indeed $30$. – Siong Thye Goh Nov 18 '18 at 04:27
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Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60 = 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

helloworld
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