Probabilistic Bounds

This chapter introduces fundamental probabilistic bounds, specifically Markov's and Chebyshev's inequalities. These powerful tools are crucial for deriving concentration inequalities and understanding the behavior of random variables without full knowledge of their distributions. Mastering these concepts is essential for advanced probability theory and frequently tested in CMI examinations.

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Chapter Contents

| Topic |

|---|-------| | 1 | Markov's Inequality | | 2 | Chebyshev's Inequality |

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We begin with Markov's Inequality.

Part 1: Markov's Inequality

Markov's Inequality provides a fundamental upper bound on the probability that a non-negative random variable takes a value greater than or equal to some positive constant, given only its expectation. This inequality is crucial for estimating probabilities when detailed distributional information is unavailable.

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Core Concepts

1. Markov's Inequality Statement

For a non-negative random variable $X$ and any positive constant $a > 0$ , Markov's Inequality states that the probability $P(X \ge a)$ is bounded by the ratio of the expected value of $X$ to $a$ .

📐 Markov's Inequality

P(X \ge a) \le \frac{E[X]}{a}

Where:
$X$ is a non-negative random variable (i.e., $P(X < 0) = 0$ ).
$E[X]$ is the expected value of $X$ .
* $a$ is a positive constant ( $a > 0$ ).
When to use: When we need an upper bound on $P(X \ge a)$ and only $E[X]$ is known, and $X$ is non-negative.

Worked Example 1: Basic Application

A random variable $X$ represents the daily energy consumption (in kWh) of a household. We know that the average daily energy consumption is $E[X] = 12$ kWh. We want to find an upper bound for the probability that the household consumes at least 30 kWh on a given day.

Step 1: Identify the random variable, its expectation, and the constant $a$ .

> Here, $X$ is the daily energy consumption, which is non-negative.
> $E[X] = 12$ .
> We are interested in $P(X \ge 30)$ , so $a = 30$ .

Step 2: Apply Markov's Inequality.

$ P(X \ge a) \le \frac{E[X]}{a}

>

P(X \ge 30) \le \frac{12}{30}

>

P(X \ge 30) \le \frac{2}{5}

>

P(X \ge 30) \le 0.4

X

P(X \ge a) \le \frac{E[X]}{a}

>

P(X \ge 2000) \le \frac{500}{2000}

>

P(X \ge 2000) \le \frac{1}{4}

>

P(X \ge 2000) \le 0.25

Y

P(Y \ge 200) \le \frac{E[Y]}{200} = \frac{100}{200} = 0.5

is not guaranteed to be non-negative. If

P(X \ge 25) \le \frac{E[X]}{25}

>

P(X \ge 25) \le \frac{5}{25}

>

P(X \ge 25) \le \frac{1}{5}

>

P(X \ge 25) \le 0.2

0.2 \times 2000

P(X \ge a) \le \frac{E[X]}{a}

>

P(X \ge 60) \le \frac{15}{60}

>

P(X \ge 60) \le \frac{1}{4}

>

P(X \ge 60) \le 0.25

0.25 \times 1000

P(X \ge a) \le \frac{E[X]}{a}

P(X \ge 450) \le \frac{150}{450}

P(X \ge 450) \le \frac{1}{3}

P(X \ge 450) \approx 0.333

X

P(X \ge 40) \le \frac{E[X]}{40}

P(X \ge 40) \le \frac{10}{40}

P(X \ge 40) \le \frac{1}{4}

P(X \ge 40) \le 0.25

0.25 \times 50 = 12.5

P(X \ge 200) \le \frac{E[X]}{200}

P(X \ge 200) \le \frac{800}{200}

P(X \ge 200) \le 4

P(X \ge 200) \le 4

P(X \ge a) \le \frac{E[X]}{a}

P(X \ge 25) \le \frac{10}{25}

P(X \ge 25) \le \frac{2}{5}

P(X \ge 25) \le 0.4

X

P(|X - 10| > 6) \le \frac{1}{k^2}

>

P(|X - 10| > 6) \le \frac{1}{3^2}

>

P(|X - 10| > 6) \le \frac{1}{9}

is outside the interval

P(|Y - \mu| > k\sigma) \le \frac{1}{k^2}

>

P(|Y - 50| > 20) \le \frac{1}{4^2}

>

P(|Y - 50| > 20) \le \frac{1}{16}

1/16

P(S - \mu \ge k\sigma) \le \frac{1}{1 + k^2}

>

P(S - 100 \ge 2 \cdot 10) \le \frac{1}{1 + 2^2}

>

P(S \ge 120) \le \frac{1}{1 + 4}

>

P(S \ge 120) \le \frac{1}{5}

120 is at most

P(R - \mu \le -k\sigma) \le \frac{1}{1 + k^2}

>

P(R - 75 \le -2 \cdot 6) \le \frac{1}{1 + 2^2}

>

P(R \le 63) \le \frac{1}{1 + 4}

>

P(R \le 63) \le \frac{1}{5} = 0.2

k=2

P(|\bar{X}_n - \mu| \ge \epsilon) \le \frac{\sigma^2}{n\epsilon^2}

\frac{\sigma^2}{n\epsilon^2} \le 0.1

\frac{25}{n \cdot 5^2} \le 0.1

>

\frac{25}{n \cdot 25} \le 0.1

>

\frac{1}{n} \le 0.1

n

n \ge \frac{1}{0.1}

>

n \ge 10

n=100

P(|\bar{X}_n - \mu| \ge \epsilon) \le \frac{\sigma^2}{n\epsilon^2}

> Substitute the values:>

P(|\bar{X}_{100} - 500| \ge 5) \le \frac{400}{100 \cdot 5^2}

>

P(|\bar{X}_{100} - 500| \ge 5) \le \frac{400}{100 \cdot 25}

>

P(|\bar{X}_{100} - 500| \ge 5) \le \frac{400}{2500}

>

P(|\bar{X}_{100} - 500| \ge 5) \le \frac{4}{25}

>

P(|\bar{X}_{100} - 500| \ge 5) \le 0.16

0.16

E[S] = E\left[\sum_{i=1}^{10} X_i\right] = \sum_{i=1}^{10} E[X_i] = \sum_{i=1}^{10} 10 = 10 \cdot 10 = 100

>

\operatorname{Var}(S) = \operatorname{Var}\left(\sum_{i=1}^{10} X_i\right) = \sum_{i=1}^{10} \operatorname{Var}(X_i) = \sum_{i=1}^{10} i

\frac{10(10+1)}{2} = \frac{10 \cdot 11}{2} = 55

P(|S - \mu_S| \ge k\sigma_S) \le \frac{1}{k^2}

>

P(|S - 100| > 15) \le \frac{1}{(15/\sqrt{55})^2}

>

P(|S - 100| > 15) \le \frac{1}{225/55}

>

P(|S - 100| > 15) \le \frac{55}{225}

>

P(|S - 100| > 15) \le \frac{11}{45} \approx 0.244

deviates from its mean by more than 15 hours is at most

E[T] = E\left[\sum_{i=1}^{20} T_i\right] = \sum_{i=1}^{20} E[T_i] = \sum_{i=1}^{20} 2 = 20 \cdot 2 = 40

T

\operatorname{Var}(T) = \operatorname{Var}\left(\sum_{i=1}^{20} T_i\right) = \sum_{i=1}^{20} \operatorname{Var}(T_i) = \sum_{i=1}^{20} (0.5 + 0.1i)

>

\operatorname{Var}(T) = \sum_{i=1}^{20} 0.5 + 0.1 \sum_{i=1}^{20} i

\frac{20(20+1)}{2} = \frac{20 \cdot 21}{2} = 210

\operatorname{Var}(T) = (20 \cdot 0.5) + (0.1 \cdot 210) = 10 + 21 = 31

and

P(|T - \mu_T| \ge M) \le \frac{\sigma_T^2}{M^2}

\frac{\sigma_T^2}{M^2} \le 0.05

\frac{31}{M^2} \le 0.05

>

M^2 \ge \frac{31}{0.05}

>

M^2 \ge 620

M

M \ge \sqrt{620}

>

M \ge 24.899...

must be an integer, the smallest integer

P(|X - \mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2}

>

P(|X - 50| > 10) \le \frac{25}{10^2}

>

P(|X - 50| > 10) \le \frac{25}{100}

>

P(|X - 50| > 10) \le 0.25

k = \epsilon/\sigma = 10/\sqrt{25} = 10/5 = 2

P(|X - \mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2}

>

P(|X - 50| > 25) \le \frac{25}{25^2}

>

P(|X - 50| > 25) \le \frac{25}{625}

>

P(|X - 50| > 25) \le \frac{1}{25}

>

P(|X - 50| > 25) \le 0.04

0.04

P(X - \mu \ge k\sigma) \le \frac{1}{1 + k^2}

>

P(X \ge 30) \le \frac{1}{1 + (2.5)^2}

>

P(X \ge 30) \le \frac{1}{1 + 6.25}

>

P(X \ge 30) \le \frac{1}{7.25}

>

P(X \ge 30) \le \frac{4}{29} \approx 0.1379

is bounded by

P(X - \mu \ge k\sigma) \le \frac{1}{1 + k^2}

>

P(X \ge 28) \le \frac{1}{1 + 2^2}

>

P(X \ge 28) \le \frac{1}{1 + 4}

>

P(X \ge 28) \le \frac{1}{5}

>

P(X \ge 28) \le 0.2$$
The numerical upper bound is 0.2."
:::

:::question type="MSQ" question="Which of the following statements are always true for a random variable $X$ with mean $\mu$ and variance $\sigma^2$ ?" options=[" $P(|X - \mu| < 2\sigma) > 0.75$ "," $P(X \ge \mu + 3\sigma) \le 1/10$ "," $P(|X - \mu| \ge 4\sigma) \le 1/16$ "," $P(X \le \mu - \sigma) \le 0.5$ "] answer=" $P(|X - \mu| < 2\sigma) > 0.75$ , $P(X \ge \mu + 3\sigma) \le 1/10$ , $P(|X - \mu| \ge 4\sigma) \le 1/16$ " hint="Apply Chebyshev's and Cantelli's inequalities to each option." solution="Let's evaluate each option:

Option 1: $P(|X - \mu| < 2\sigma) > 0.75$
> By Chebyshev's inequality, $P(|X - \mu| \ge k\sigma) \le 1/k^2$ .
> For $k=2$ , $P(|X - \mu| \ge 2\sigma) \le 1/2^2 = 1/4 = 0.25$ .
> This means $1 - P(|X - \mu| < 2\sigma) \le 0.25$ .
> So, $P(|X - \mu| < 2\sigma) \ge 1 - 0.25 = 0.75$ .
> The statement $P(|X - \mu| < 2\sigma) > 0.75$ is true (it's $\ge 0.75$ ).

Option 2: $P(X \ge \mu + 3\sigma) \le 1/10$
> This is a one-sided bound. Using Cantelli's inequality:
> $P(X - \mu \ge k\sigma) \le 1/(1+k^2)$ .
> For $k=3$ , $P(X - \mu \ge 3\sigma) \le 1/(1+3^2) = 1/(1+9) = 1/10$ .
> The statement is true.

Option 3: $P(|X - \mu| \ge 4\sigma) \le 1/16$
> By Chebyshev's inequality, $P(|X - \mu| \ge k\sigma) \le 1/k^2$ .
> For $k=4$ , $P(|X - \mu| \ge 4\sigma) \le 1/4^2 = 1/16$ .
> The statement is true.

Option 4: $P(X \le \mu - \sigma) \le 0.5$
> This is a one-sided bound. Using Cantelli's inequality:
> $P(X - \mu \le -k\sigma) \le 1/(1+k^2)$ .
> For $k=1$ , $P(X - \mu \le -\sigma) \le 1/(1+1^2) = 1/2 = 0.5$ .
> The statement is true.

Wait, all options are true. An MSQ should have multiple correct options, but typically not all.
Let me re-check the problem statement again. "Which of the following statements are always true".
All of them seem always true based on Chebyshev's and Cantelli's.
If it's an MSQ, I need to ensure it's not all true, or that some are incorrect.
Let's try to make one false.
If I make Option 4 $P(X \le \mu - \sigma) < 0.5$ , it would be false because it can be exactly 0.5.
Let's make Option 1 $P(|X - \mu| < 2\sigma) < 0.75$ . This would be false.

I will modify Option 4 to be false.
New Option 4: " $P(X \le \mu - \sigma) < 0.5$ "
Solution for New Option 4:
> $P(X \le \mu - \sigma) \le 0.5$ is true by Cantelli's. However, the inequality is $\le$ , not $<$ . It is possible for the bound to be exactly $0.5$ (e.g., for a distribution concentrated at the mean). Therefore, $P(X \le \mu - \sigma) < 0.5$ is not always true. It could be $0.5$ .
So, this makes Option 4 false.

Let's ensure the solution explains why it's not always true.
"Option 4: $P(X \le \mu - \sigma) < 0.5$
> By Cantelli's inequality, $P(X - \mu \le -k\sigma) \le 1/(1+k^2)$ .
> For $k=1$ , $P(X - \mu \le -\sigma) \le 1/(1+1^2) = 1/2 = 0.5$ .
> This means the probability is at most $0.5$ . It can be equal to $0.5$ for some distributions. For example, consider a degenerate distribution where $X = \mu - \sigma$ with probability $0.5$ and $X = \mu + \sigma$ with probability $0.5$ . In this case, $\operatorname{Var}(X) = (\mu-\sigma-\mu)^2 \cdot 0.5 + (\mu+\sigma-\mu)^2 \cdot 0.5 = \sigma^2 \cdot 0.5 + \sigma^2 \cdot 0.5 = \sigma^2$ . So this is a valid distribution. For this distribution, $P(X \le \mu - \sigma) = 0.5$ .
> Thus, the strict inequality $P(X \le \mu - \sigma) < 0.5$ is not always true. It can be equal to $0.5$ .
> This statement is false."
:::

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Summary

❗ Key Formulas & Takeaways

| Formula/Concept | Expression |

|---|----------------|------------| | 1 | Chebyshev's Inequality (General) |

P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}

P(|X - \mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2}

| | 2 | Cantelli's Inequality (One-Sided) |

P(X - \mu \ge k\sigma) \le \frac{1}{1 + k^2}

and

P(X - \mu \le -k\sigma) \le \frac{1}{1 + k^2}

| | 3 | Chebyshev's for Sample Mean |

P(|\bar{X}_n - \mu| \ge \epsilon) \le \frac{\sigma^2}{n\epsilon^2}

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What's Next?

💡 Continue Learning

This topic connects to:

Law of Large Numbers: Chebyshev's inequality is fundamental to proving the Weak Law of Large Numbers, showing that the sample mean converges in probability to the true mean.

Concentration Inequalities: It is a basic example of a concentration inequality, which bound the probability that a random variable takes on a value far from its expected value. More advanced inequalities like Chernoff bounds offer tighter bounds for sums of independent random variables, especially for specific distributions.

Chapter Summary

❗ Probabilistic Bounds — Key Points

Markov's Inequality: For a non-negative random variable $X$ and any $a > 0$ , it provides an upper bound for the probability $P(X \ge a) \le \frac{E[X]}{a}$ . It is fundamental but often provides loose bounds.

Chebyshev's Inequality: For any random variable $X$ with finite mean $\mu$ and finite variance $\sigma^2$ , it bounds the probability of deviation from the mean: $P(|X - \mu| \ge k) \le \frac{\sigma^2}{k^2}$ for any $k > 0$ .

Relationship: Chebyshev's Inequality can be directly derived by applying Markov's Inequality to the non-negative random variable $(X - \mu)^2$ .

Universality: Both inequalities are powerful because they provide probability bounds without requiring knowledge of the full probability distribution of the random variable.

Conditions: Markov's requires the random variable to be non-negative. Chebyshev's requires the existence of a finite mean and variance.

Utility: They are crucial for theoretical proofs (e.g., the Weak Law of Large Numbers) and for practical quick estimates of probabilities when detailed distributional information is unavailable.

Limitations: While universally applicable under their conditions, the bounds provided by these inequalities are often not tight, meaning the actual probability might be much smaller than the upper bound.

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Chapter Review Questions

:::question type="MCQ" question="A non-negative random variable $X$ represents the time (in minutes) a customer waits in a queue. If the expected waiting time $E[X]$ is $15$ minutes, what is the maximum probability that a customer will wait for $30$ minutes or more?" options=["0.25","0.5","0.75","1.0"] answer="0.5" hint="Apply Markov's Inequality directly." solution="Using Markov's Inequality, $P(X \ge a) \le \frac{E[X]}{a}$ . Here, $E[X] = 15$ and $a = 30$ .
$P(X \ge 30) \le \frac{15}{30} = 0.5$ .
Thus, the maximum probability is $0.5$ ."
:::

:::question type="NAT" question="A stock's daily return $R$ has a mean of $0.002$ and a variance of $0.0004$ . Using Chebyshev's Inequality, find an upper bound for the probability that the absolute deviation of the return from its mean is at least $0.05$ . Round your answer to two decimal places." answer="0.16" hint="Recall Chebyshev's inequality: $P(|X - \mu| \ge k) \le \frac{\operatorname{Var}(X)}{k^2}$ ." solution="Given $\mu = 0.002$ , $\operatorname{Var}(R) = 0.0004$ , and $k = 0.05$ .
Applying Chebyshev's Inequality:
$P(|R - 0.002| \ge 0.05) \le \frac{0.0004}{(0.05)^2} = \frac{0.0004}{0.0025} = \frac{4}{25} = 0.16$ .
The upper bound for the probability is $0.16$ ."
:::

:::question type="MCQ" question="Which of the following statements regarding probabilistic bounds is TRUE?" options=["Markov's Inequality can be applied to any random variable, positive or negative." , "Chebyshev's Inequality requires the random variable to have a symmetric distribution." , "Chebyshev's Inequality provides a bound using only the mean, while Markov's uses mean and variance." , "Chebyshev's Inequality can be derived from Markov's Inequality by considering the squared deviation from the mean."] answer="Chebyshev's Inequality can be derived from Markov's Inequality by considering the squared deviation from the mean." hint="Review the fundamental conditions and derivations of each inequality." solution="Let's evaluate each option:
* 'Markov's Inequality can be applied to any random variable, positive or negative.' - False. Markov's Inequality specifically requires the random variable to be non-negative.
* 'Chebyshev's Inequality requires the random variable to have a symmetric distribution.' - False. Chebyshev's Inequality applies to any distribution with finite mean and variance; symmetry is not a requirement.
* 'Chebyshev's Inequality provides a bound using only the mean, while Markov's uses mean and variance.' - False. Markov's uses only the mean. Chebyshev's uses both the mean (as a reference point for deviation) and the variance.
* 'Chebyshev's Inequality can be derived from Markov's Inequality by considering the squared deviation from the mean.' - True. Chebyshev's Inequality is derived by applying Markov's Inequality to the non-negative random variable $(X - \mu)^2$ ."
:::

:::question type="NAT" question="A machine produces components with a mean length of $100$ mm and a standard deviation of $2$ mm. Using Chebyshev's Inequality, what is the minimum proportion of components that must fall within the length range of $96$ mm to $104$ mm? Express your answer as a decimal rounded to two places." answer="0.75" hint="First, find the upper bound for the proportion of components outside the given range using Chebyshev's Inequality. Then subtract this from $1$ to find the minimum proportion within the range." solution="Given $\mu = 100$ mm and $\sigma = 2$ mm. Thus, $\operatorname{Var}(X) = \sigma^2 = 2^2 = 4$ .
The length range is $96$ mm to $104$ mm. This means a deviation of $k = 104 - 100 = 4$ mm from the mean.
Using Chebyshev's Inequality, the upper bound for the probability of a component falling outside this range is:
$P(|X - \mu| \ge k) \le \frac{\operatorname{Var}(X)}{k^2} = \frac{4}{4^2} = \frac{4}{16} = 0.25$ .
This means at most $25\%$ of components can fall outside the range.
Therefore, the minimum proportion of components that must fall within the range is $1 - 0.25 = 0.75$ .
The answer is $0.75$ ."
:::

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What's Next?

💡 Continue Your CMI Journey

Having understood probabilistic bounds, you are now well-equipped to explore their profound implications in probability theory. The next logical step is to delve into Laws of Large Numbers, particularly the Weak Law of Large Numbers (WLLN), which directly leverages Chebyshev's Inequality to demonstrate how sample means converge to the true population mean as the sample size increases. This foundational concept is critical for understanding statistical inference. Following this, you can proceed to the Central Limit Theorem (CLT), which provides a more specific and powerful approximation for the distribution of sample means, further solidifying your understanding of statistical estimation and hypothesis testing.

Probabilistic Bounds

Probabilistic Bounds

Chapter Contents

| Topic |

Part 1: Markov's Inequality

Core Concepts

1. Markov's Inequality Statement

2. Conditions and Limitations

Advanced Applications

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Part 2: Chebyshev's Inequality

Core Concepts

1. Chebyshev's Inequality (General Form)

2. One-Sided Chebyshev's Inequality (Cantelli's Inequality)

3. Application to Sample Mean

Advanced Applications

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

| Formula/Concept | Expression |

What's Next?

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Probability Theory

Basic Probability

Random Variables

Expectation and Variance

Conditional Probability and Independence

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers

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Chapter Practice

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