Chebyshev Sum Inequality

Another important algebraic inequality is the Chebyshev sum inequality which states that for similarly sorted sequences of real numbers - both monotonically decreasing: a, > a₂ >,...,> a„ and (?, > b₂ >,...,> b,„ or both monotonically increasing: a, < a₂ <,...,< я„; b2 <,...,< b,„ the following sharp inequality holds:

or

If the sequences are oppositely sorted (for example, a, >,...,> a„, b <,...,< b„ or a, < a₂ <,...,< a,„ b > b₂ b„) the inequality is reversed.

Equality is attained if a_t =a₂=.... = a„ or b, = b₂ =.... = b„ holds.

Despite the existence of a number of alternative proofs of the Chebyshev’s sum inequality (Besenyei, 2018), probably the simplest proof is based on the rearrangement inequality.

According to the rearrangement inequality, for similarly ordered sequences, the following inequalities are true:

By adding these inequalities, the inequality

is obtained, which, after taking out the common factor (/?, + b₂ +... + b„), leads to the Chebyshev inequality (2.36).

Chebyshev inequality provides the valuable opportunity to segment an initial complex expression into simpler expressions. The complex terms яД in (2.36) are segmented into simpler terms involving a, and b,. The segmentation capability of the Chebyshev inequality will be illustrated by evaluating the lower bound of X"_=]x,² if jq +x₂ + ... + л:„ = 1.

Without loss of generality, it can be assumed that x, <x₂ <...ba₂ = x₂,—,a„ = x„ and b = x,b₂ = x₂,...,b„ = x„, the conditions for applying the Chebyshev inequality are fulfilled and

holds. Substituting x, +... + x„ = 1 in (2.38), gives the lower bound of £"₌₁xf:

Muirhead Inequality

Consider the two non-increasing sequences а_л > a₂ >,...,> a„ and b_t > b₂ >,...,> b„ of positive real numbers. The sequence {a) is said to majorise the sequence {/?) if the following conditions are fulfilled:

If the sequence |a) majorises the sequence [b and x_bx₂,...,x„ are non-negative, the Muirhead inequality

holds (Hardy et al., 1999).

For any set of non-negative numbers X],x₂,...,x„, the symmetric sum X^x^xf—x“", when expanded, includes n terms. Each term of the symmetric sum is formed by a distinct permutation of the elements of the sequence a_ua₂,...,a„. Thus, if {a} = [2,1,0], the symmetric sum becomes

If (a) = [2,0,0], the symmetric sum becomes

Here is an application example featuring an inequality that follows directly from the Muirhead inequality. Consider a set of real, non-negative numbers x,,x₂,x₃. It can be shown that the following inequality holds:

Consider the set of non-negative numbers x,, x₂, xj and the sequences (a) = [2,0,0] and {b) =[1,1,0]. Clearly, the sequence {a} = [2,0,0] majorises the sequence {b) = [1,1,0] because the conditions (2.39) are fulfilled:

According to the Muirhead inequality (2.40):

which implies inequality (2.41).

Markov Inequality

Markov’s inequality is related to a random variable X with mean ц = E(X), which accepts only non-negative values. It states that the probability that the random variable will be greater than a positive constant a does not exceed the value /.i/a:

This inequality is very general. It holds for any distribution of the random variable X.

Proof. For a continuous probability density distribution f(x) of the random variable X, the expected value is given by

Because, the random variable X accepts only non-negative values, all the terms in the right side of (2.43) are non-negative. Dropping the first term in equation (2.43) and replacing ‘x’ in the second term with the smallest value ‘a’ yields

Since P(X >a)= f f(x)dx. finally, inequality (2.44) yields

xta

which is the Markov inequality (2.42). Replacing the integrals with sums provides a proof of the Markov inequality for a random variable X with a discrete distribution.

Markov’s inequality provides a useful tight upper bound of the values of a random variable for large values of the constant ‘a’. Thus, if the mean of the random variable describing loss given failure is ju, according to the Markov inequality, the maximum possible value of the probability that the loss will exceed for example, the value lO^u is

This is a tight upper bound, which is actually attained if P(X <10ju) = 0 and P{X =10p) = l-

Chebyshev Inequality

Consider a random variable X with mean ц and variance У(Х). For any constant к > 0, the Chebyshev inequality states that

Considering that V(X) = o¹, setting к = to where t > 0 in (2.45) gives

The last formulation of the Chebyshev inequality states that for any positive constant t, the probability that the random variable X will take on a value more than t standard deviations from the mean ц = E(X) does not exceed 1/r.

The Chebyshev inequality is also very general. It holds for any distribution of the random variable X.

Proof. Consider the auxiliary non-negative random variable Y = [X - ц]¹. Because the random variable Y is non-negative, the Markov inequality (2.42) can be applied and

Considering that E(Y) = E[(X - ц)²] = V(X) and P(Y > k²) = P(X-pi > k), the Chebyshev inequality (2.45) is finally obtained.

Chebyshev’s inequality also provides a useful tight upper bound of the values of a random variable for values of the constant к greater than the standard deviation a. The application of the inequality will be illustrated by an example.

Consider loss given failure with mean /и. Suppose that the extreme value of the loss which triggers bankruptcy is 5cr, where a is the standard deviation of the loss. According to the Chebyshev inequality, the probability that the loss will be larger than the extreme value of 5cr does not exceed 0.04:

The probability value of 0.04 is a tight upper bound for the extreme loss, which holds irrespective of the actual distribution of the loss given failure.

Minkowski Inequality

The general Minkowski inequality for two sequences a_l,a₂,...,a„ and b_ub₂,...,b_n of non-negative numbers states (Hardy et al, 1999)

for p > 1.

For p < 1, the inequality direction is reversed (Bechenbach and Bellman, 1961). For /7 = 2, the general triangle inequality is obtained:

It is not difficult to see that the general triangle inequality is valid for arbitrary sequences a_{,a₂,...,a„ and b_ub₂,...,b_n of real numbers (not necessarily non-negative).

If n = 3, from inequality (2.48) the three-dimensional triangle inequality is obtained

which expresses geometrically that the sum of the two sides of the triangle with vertices (0, 0, 0), (я,, a₂, a_}) and (a, + b_ha₂ + Ь₂,а_ъ + /?,) is greater than or equal to the length of the third side (Figure 2.2).

If и = 2, from inequality (2.48) the two-dimensional triangle inequality is obtained

which expresses geometrically that the sum of the two sides of a triangle is greater than or equal to the length of the third side (Figure 2.3).

FIGURE 2.2 Geometric interpretation of the three-dimensional triangle inequality.

FIGURE 2.3 Geometric interpretation of the two-dimensional triangle inequality.

If n = 1, for arbitrary real numbers а_л and b_t, from the general triangle inequality (2.48) the inequality

is obtained, which is equivalent to

which is the one-dimensional triangle inequality (2.7).