Математика. Материалы курса

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The characteristic function of a random variable is a function of the real parameter t, which is calculated as the mathematical expectation of the following function е^itξ: ψ(t)=M[е^itξ], where ξ is a random variable, and i is an imaginary unit.

Thus, the values of the characteristic function are complex, that is, the function ψ: R→C acts from the set of real numbers to the set of complex numbers.

Let's learn how to search for characteristic functions for discrete and continuous distributions.

Example 1. A random variable ξ is an indicator of a random event (i.e. it has a Bernoulli distribution) with the probability p. We should construct a characteristic function of this random variable.

Let ξ be an indicator of a random event, that is, the distribution of the value ξ is discrete and has the following form: the value 0 and the value 1 are the set of values of the random variable. It takes the value 1 with the probability p, and 0 with the probability 1-p (see the video).

Then the characteristic function of this random variable at the point t is equal to the mathematical expectation of the discrete random variable. To calculate it, we need to multiply the probability of a random variable by its value. The random variable е^itξ takes the value 1 and the value е^it. Thus, we get: ψ(t)= е^it0(1–p)+е^it1p. е^it0=1. This means that we get 1–p+е^itp.

If we consider complex random variables, their characteristic functions look more complex as well (there are more summands, which may be reduced). Thus, we can find a characteristic function for any random variable that has a discrete distribution.

Let us find a characteristic function for a random variable with a continuous distribution.

We are going to consider an equiprobability distribution on the segment [a, b].

Example 2. The random variable ξ has a equiprobability distribution on the segment [a, b]. We have to construct the characteristic function of this random variable.

The distribution density is a constant for this random variable on the segment [a, b]: f(x)_ξ= 1/(b–a) if a≤x≤b, and 0 if x<a or x>b (see the video). Then the characteristic function of the value ξ at the point t ψ (t) is equal to the integral from –∞ to +∞ of the product, and the second factor is the distribution density.

Since the distribution density differs from zero only on the interval from a to b, we can count the integral only on the interval from a to b here.

ψ(t)= see the video. 1/(b–a) is a constant, it can be taken out of the integral sign. The exponent remains under the integral. We know that the integral of the exponent is the same exponent, but we need to divide by the coefficient at x. Therefore, this integral is equal to е^itx/(it(b–a)) (see the video). We should consider this in the range from a to b. That is, instead of x, we substitute first b, then a.

If we work with complex-valued functions, we do not leave the imaginary unit in the denominator, so we need to multiply the numerator and denominator by –i, then there is no imaginary unit in the denominator.

Get –1/(t(b–a))(е^itb–е^ita) (see video). In other words, the characteristic function for an equiprobability distribution has this form.

In the second lecture, we are going to use this fact to understand whether an equiprobability distribution is stable.

In the practical class, you also need to find characteristic functions for some distributions. May success attend you!