# What are some intuitive proofs of Markov’s inequality and Chebyshev’s inequality?

Markov's inequality: Suppose that an average human is 6 feet tall. Then the people who are 60 or more feet tall form at most 10 percent of the population.
Proof: The premise implies that if the total number of humans is n, their total height in feet is 6n. If you had more than n/10 people who are each taller than 60 feet, already the sum of their heights (ignoring the other 9n/10 people) would exceed 6n.
Note that the above proof uses an important silent assumption that the height of each person is nonnegative.
In general, Markov's inequality tells you that a nonnegative random variable must behave nicely: it cannot frequently exceed its mean by a large value. From the above visualisation it may seem useless, but that's not true: the more useful Chebyshev's inequality is its consequence.
For Chebyshev's inequality we can use any random variable (i.e., it doesn't have to be nonnegative any more). This inequality uses not just the mean but also another statistic: the standard deviation.
Chebyshev's inequality: As above, assume that the mean human height is 6 feet. Additionally, let's assume that the stdev is 4 inches = 1/3 of a foot. Then at least 8/9 of all humans must have between 5 and 7 feet.
Proof: Suppose we pick a random human and compute the difference between their height and the ideal height of 6 feet. The square of this difference is a nonnegative random variable. Let's call it Y. The mean of Y is σ 2  σ2\sigma^2, where σ σ\sigma is our standard deviation (4 inches). By Markov's inequality, at most 1/9 of all humans can have Y greater than 9σ 2 =(3σ) 2  9σ2=(3σ)29\sigma^2 = (3\sigma)^2. In other words, at most 1/9 of all humans can differ from the mean height by 3σ= 3σ=3\sigma= 1 foot or more.
In general, Chebyshev's inequality tells you that the probability that a random variable falls at least kσ kσk\sigma away from the mean is at most 1/k 2  1/k21/k^2. This must be true for any random variable, regardless of its probability distribution.

Markov's inequality: Suppose that an average human is 6 feet tall. Then the people who are 60 or more feet tall form at most 10 percent of the population.
Proof: The premise implies that if the total number of humans is n, their total height in feet is 6n. If you had more than n/10 people who are each taller than 60 feet, already the sum of their heights (ignoring the other 9n/10 people) would exceed 6n.
Note that the above proof uses an important silent assumption that the height of each person is nonnegative.
In general, Markov's inequality tells you that a nonnegative random variable must behave nicely: it cannot frequently exceed its mean by a large value. From the above visualisation it may seem useless, but that's not true: the more useful Chebyshev's inequality is its consequence.
For Chebyshev's inequality we can use any random variable (i.e., it doesn't have to be nonnegative any more). This inequality uses not just the mean but also another statistic: the standard deviation.
Chebyshev's inequality: As above, assume that the mean human height is 6 feet. Additionally, let's assume that the stdev is 4 inches = 1/3 of a foot. Then at least 8/9 of all humans must have between 5 and 7 feet.
Proof: Suppose we pick a random human and compute the difference between their height and the ideal height of 6 feet. The square of this difference is a nonnegative random variable. Let's call it Y. The mean of Y is $\sigma^2$, where $\sigma$ is our standard deviation (4 inches). By Markov's inequality, at most 1/9 of all humans can have Y greater than $9\sigma^2 = (3\sigma)^2$. In other words, at most 1/9 of all humans can differ from the mean height by $3\sigma=$ 1 foot or more.
In general, Chebyshev's inequality tells you that the probability that a random variable falls at least $k\sigma$ away from the mean is at most $1/k^2$. This must be true for any random variable, regardless of its probability distribution.

What are some intuitive proofs of Markov's inequality and Chebyshev's inequality?