[Math] How to use Chebyshev’s inequality for this problem

chebyshev-functionstatistics

A probability distribution has a mean of 50 and a standard deviation of $2$. Use Chebyshev's inequality to find the minimum probability that an outcome is between $42$ and $58$.

How would I use Chebyshev's inequality for this problem? Any similar questions, feel free to link it in the comment section, as well as any questions. Thanks

Best Answer

A big hint:

$$42 < X < 58$$

is the same as

$$|X-50|<8$$

Related Solutions

[Math] Chebyshev’s inequality for 1 standard deviation results in 0

Chebyshev's Inequality is true for any $c > 0$, but you are right that it only provides useful information for $c > 1$.

This is actually surprisingly easy prove. Define $\mu = E(X)$ and $\sigma^2 = E((X-\mu)^2)$. Observe that for any $c \geq 0$ we have $\mathbb{1}\left\{\left|\frac{X-\mu}{\sigma}\right|\geq c\right\} \leq \frac{(X-\mu)^2}{\sigma^2 c^2}$ where $\mathbb{1}\{\cdot\}$ is an indicator function equal to 1 if the event inside the brackets occurs and 0 otherwise. (To see this, suppose $\left|\frac{X-\mu}{\sigma}\right| \geq c$. Then the left-hand side equals 1 but the right-hand side is $\geq$ 1. If $\left|\frac{X-\mu}{\sigma}\right| < c$, the left-hand side is 0 but the right-hand side is positive.)

Since the expectation operator is monotonic, we can take expectations of both sides to obtain $$ \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) \leq \frac{E(X-\mu)^2}{\sigma^2 c^2} = \frac{1}{c^2} $$ Since $\Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) = 1- \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\leq c\right)= 1 -\Pr(\mu- \sigma c \leq X \leq \mu + \sigma c)$, we can rearrange to obtain the desired inequality.

The proof illustrates how you could derive many other inequalities of this type. Any function $g$ that satisfies

$$\left|\frac{X-\mu}{c\sigma}\right|\geq 1 \Rightarrow g\left(\left|\frac{X-\mu}{c\sigma}\right|\right) \geq 1$$

will work. For example, you could use an identical proof to show $$ \Pr\left(\left|\frac{X-\mu}{\sigma}\right|\geq c\right) \leq \frac{E\left(|X-\mu|^k\right)}{\sigma^k c^k} $$ for all $k > 0$.

[Math] How to round a discrete random variable

In a practical situation, in order to formulate the limits of probabilities such as $P(7.22 < X < 15.99)$ for a random variable $X$ that takes only integer values, you would need a good reason for picking the exact endpoints 7.22 and 15.99. (Why not endpoints 7.23 and 16.00?)

However, once formulated, there can be no question that $$P(7.22 < X < 15.99) = P(7.22 \le X \le 15.99) = \sum_{i=8}^{15} P(X = i).$$ That is, the sum of eight individual probabilities. For example, the binomial random variable $X \sim Binom(20, 1/2)$ has positive probability only at integer values. It makes no difference what non-integer values are included between the stated endpoints. In particular, for this binomial random variable, we have $P(7.22 < X < 15.99) = 0.862503,$ computed in R as follows.

 i = 8:15;  sum(dbinom(i, 20, 1/2))
 ## 0.862503

Of course, one could also use the binomial PDF formula to compute each of the eight terms, and then add them together.

I hope it is now clear that $P(7.22 < X < 16.00)$ and $P(7.22 \le X \le 16.00)$ have different values, unless it happens that $P(X = 16) = 0.$

Best Answer

Related Solutions

[Math] Chebyshev’s inequality for 1 standard deviation results in 0

[Math] How to round a discrete random variable

Related Question