Chebyshev’s Inequality, and an Application
Chebyshev’s Inequality: Let Y be a random variable whose distribution has mean µ and
standard deviation σ < +∞. Then for any k ≥ 1, we have
1
1
𝑃(|𝑌 − 𝜇| ≥ 𝑘𝜎) ≤ 𝑘 2 or, equivalently, 𝑃(|𝑌 − 𝜇| ≤ 𝑘𝜎) ≥ 1 − 𝑘 2 .
Proof: (Continuous case; the proof for the discrete case is parallel.) By definition,
+∞
2
𝜎 = 𝐸[(𝑌 − 𝜇)
2]
=∫
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦.
−∞
The interval of integration may be partitioned as follows:
𝜇−𝑘𝜎
𝜎2 = ∫
𝜇+𝑘𝜎
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦 + ∫
−∞
+∞
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦 + ∫
𝜇−𝑘𝜎
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦.
𝜇+𝑘𝜎
Now in each of the integrals the integrand is non-negative everywhere (except perhaps at a
countable number of points). Hence, if we drop the second term on the right, we obtain
𝜇−𝑘𝜎
𝜎2 ≥ ∫
+∞
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦 + ∫
−∞
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦.
𝜇+𝑘𝜎
Now, on the interval (−∞, 𝜇 − 𝑘𝜎), we have |𝑦 − 𝜇| ≥ 𝑘𝜎. Also, on the interval (−∞, 𝜇 + 𝑘𝜎),
we have |𝑦 − 𝜇| ≥ 𝑘𝜎. Hence, if we replace (𝑦 − 𝜇)2 in each of the above integrals by 𝑘𝜎, each
integral will become smaller; i.e.,
𝜇−𝑘𝜎
𝜇−𝑘𝜎
2
(𝑘𝜎)2 𝑓𝑌 (𝑦)𝑑𝑦,
(𝑦 − 𝜇) 𝑓𝑌 (𝑦)𝑑𝑦 ≥ ∫
∫
−∞
−∞
and
𝜇+𝑘𝜎
∫
𝜇+𝑘𝜎
(𝑦 − 𝜇)2 𝑓𝑌 (𝑦)𝑑𝑦 ≥ ∫
−∞
(𝑘𝜎)2 𝑓𝑌 (𝑦)𝑑𝑦.
−∞
Therefore, we have
𝜇−𝑘𝜎
𝜎2 ≥ ∫
+∞
(𝑘𝜎)2 𝑓𝑌 (𝑦)𝑑𝑦 + ∫
−∞
(𝑘𝜎)2 𝑓𝑌 (𝑦)𝑑𝑦.
𝜇+𝑘𝜎
We may then factor the constants out of each integral to obtain
𝜇−𝑘𝜎
𝜎 2 ≥ (𝑘𝜎)2 ∫
−∞
+∞
𝑓𝑌 (𝑦)𝑑𝑦 + (𝑘𝜎)2 ∫
𝜇+𝑘𝜎
𝑓𝑌 (𝑦)𝑑𝑦.
Now, the first integral in the above equation is
𝜇−𝑘𝜎
𝑓𝑌 (𝑦)𝑑𝑦 = 𝑃(𝑌 ≤ 𝜇 − 𝑘𝜎),
∫
−∞
while the second integral is
+∞
𝑓𝑌 (𝑦)𝑑𝑦 = 𝑃(𝑌 ≥ 𝜇 + 𝑘𝜎).
∫
𝜇+𝑘𝜎
Substituting these two equalities into the previous inequality, we obtain
𝜎 2 ≥ (𝑘𝜎)2 [𝑃(𝑌 ≤ 𝜇 − 𝑘𝜎) + 𝑃(𝑌 ≥ 𝜇 + 𝑘𝜎)], or 𝜎 2 ≥ 𝑘 2 𝜎 2 𝑃(|𝑌 − 𝜇| ≥ 𝑘𝜎).
If we divide through the equality by the common factor 𝑘 2 𝜎 2 , we obtain the result
𝑃(|𝑌 − 𝜇| ≥ 𝑘𝜎) ≤
1
.
𝑘2
Equivalently, by the complement rule, we may write this result as
𝑃(|𝑌 − 𝜇| < 𝑘𝜎) ≥ 1 −
1
.
𝑘2
These last two statements are equivalent forms of Chebyshev’s Inequality.
Example: The target value of the length of a type of rivet manufactured by a certain process is
50 mm. If we assume that the mean length for the population of all rivets manufactured by this
process is actually 𝜇 = 50 𝑚𝑚, then what is the maximum tolerable value for the population
standard deviation, σ, if we want the proportion of rivets with lengths between 49.99 mm and
50.01 mm to be at least 93.75%?
1
From the latter form of Chebyshev’s Inequality, we have 1 − 𝑘 2 = 0.9375, or
𝑘 = 4.
Thus, we write
𝑃(|𝑌 − 𝜇| ≤ 4𝜎) ≥ 0.9375,
or
4𝜎 = 0.01 𝑚𝑚,
or 𝜎 = 0.0025 𝑚𝑚.
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