# Use Chebychev's Inequality to find a lower bound.

Let $X$ be the number of heads one would obtain in $140$ flips of a fair coin.

Use Chebychev's Inequality to find a lower bound on the probability $P(60 < X < 80)$.

Okay so Chebychev's Inequality is $P(|X - E(X)| > kσ) \le 1/k^2$ for $k > 0$, where $σ^2$ is the variance of $X$.

I'm not sure how to fill this in or anything. My probabilty test is tomorrow so help is much appreciated! Descriptive answers would be awesome.

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Which is the $E(X)$ and $\sigma^2$ in your case? Did you compute them? – leonbloy Aug 14 '12 at 18:24

## 3 Answers

Note that $X$ is binomial with $n=140, p = 0.5$ so $\mathbb{E}[X] = np = 70$, $\sigma^2 = Var(X) = np(1-p) = 35,$ giving $\sigma = \sqrt{35}$ .

Now,

$\begin{split} \mathbb{P}[60 < X < 80] &= \mathbb{P}[-10 < X - \mathbb{E}[X] < 10] \\ &= \mathbb{P}[|X - \mathbb{E}[X]| < 10] \\ &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10}{\sqrt{35}} \sigma]\\ &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10\sigma}{\sqrt{35}}] \end{split}$

By Chebyshev's Inequality, $\mathbb{P}[X \not \in (60,80)] = \mathbb{P}[|X - \mathbb{E}[X]| > \frac{10\sigma}{\sqrt{35}}] \leq (\frac{\sqrt{35}}{10})^2 = 35/100 = 7/20$.

Hence, $\mathbb{P}[60 < X < 80] = 1 - \mathbb{P}[X \not \in (60,80)] \geq 1-7/20 = 13/20.$

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Why is p = 0.5? (Just trying to understand the question a little better as it won't be identical tomorrow!) – Fred Aug 14 '12 at 18:34
@Panda Because you said you are using a fair coin. Probability of success here (named $p$) is probability of tossing a head, which must equal the probability of tossing a tail, since the coin is fair. There are only 2 outcomes of a toss, with equal probabilities, hence each must be $1/2$. – gt6989b Aug 14 '12 at 18:38
I've gotten terrible at this stuff. Would've seen that stuff straight away before! Thanks so much! Gonna be a long night :'( – Fred Aug 14 '12 at 18:42
@gt6989b: There may be a few typos in the solution. – André Nicolas Aug 14 '12 at 22:08
What is the difference between an upper bound and a lower bound? Are they the same? I thought the OP wanted a lower bound on $P\{60 < X < 80\}$ but you have provided an upper bound instead since you get $$P\{60 < X < 80\} < \left(\frac{4}{49}\right).$$ – Dilip Sarwate Aug 15 '12 at 3:57

Before the solution, a minor comment. The Chebyshev Inequality is not quite quoted correctly. It should be $$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.\tag{1}$$ For continuous distributions there is no need to distinguish between $\le$ and $\lt$. Here we are working with a discrete distribution.

A standard calculation shows that in our case $\mu=np=70$ and $\sigma^2=np(1-p)=35$. We want a lower bound on $\Pr(60\lt X\lt 80)$. The complementary event is $|X-70|\ge 10$. We first find an upper bound for $\Pr(|X-70|\ge 10)$.

Compare with Inequality $(1)$ quoted above. In our case we have $k\sigma=10$, and therefore $$k=\frac{10}{\sigma}, \quad\text{so}\quad\frac{1}{k^2}=\frac{\sigma^2}{100}=\frac{35}{100}.$$

It follows that $\Pr(|X-70|\ge 10)$ is $\le \frac{35}{100}$. Thus $$\Pr(60\lt X\lt 80)\ge 1-\frac{35}{100}=\frac{65}{100}.$$ That is the lower bound given by the Chebyshev Inequality.

Remark: It is not a very good lower bound. You might want to use software such as the free-to-use Wolfram Alpha to calculate the exact probability. It's not Chebyshev's fault. An inequality that works for every distribution that has a mean and variance, including some pretty weird ones, cannot be expected to compete against estimates based on more information.

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Hint: You should recognize this experiment as a series of Bernoulli trials, and so its probability distribution is given by the binomial distribution! Compute the mean and variance and fill in the blanks and you are basically done!

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How do you know how to recognise questions as Binomial or Normal etc? – Fred Aug 14 '12 at 18:32
When you learn Bernoulli trials, you learn why their distributions are binomial. How did I identify it as a Bernoulli experiement then? I'm using the mathematical principle: "If it looks like a duck and quacks like a duck, it's a duck." That is to say it satisfies the definition of what a Bernoulli trial is. I think you need to reread your text section on Bernoulli trials, as well as their explanation of how the binomial distribution comes out of it. – rschwieb Aug 14 '12 at 18:36
I suppose I should add one thing to that: you probably learned that the normal distribution can approximate the binomial distribution, so you might me confused as to when you would want to use the normal distribution. It is advantageous to use the normal approximation if the number of trials is really high, or else the number of individual events you need to compute is large. – rschwieb Aug 14 '12 at 18:39
@Panda Binomial distribution is a sum of Bernoulli trials. Normal is typically modeling some continuous quantity, e.g., error of some sort. For example, amount of soup automatically dispensed into a container would vary somewhat, and it is typical to model as Normal. – gt6989b Aug 14 '12 at 18:40
@Panda The beauty of Chebyshev's Inequality is that it works for all distributions :). Anything where you can compute expected value and variance will do. – gt6989b Aug 14 '12 at 18:45