Математика. Материалы курса

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Practical task: Testing the robustness for standard distributions.

Let me remind you that a distribution is considered robust if the sum of random variables that have this distribution also has a similar distribution. That is, for example, if we are talking about a binomial distribution, if the sum of random variables that have a binomial distribution also has a binomial distribution, possibly with other parameters, we will say that the binomial distribution is robust. At the lecture, we showed that the random event indicator and uniform distribution will not be robust.

What about binomial distributions? The task is to find the characteristic function of a random variable, which is a composition of random variables that have a binomial distribution, and answer the question if it is robust or not.

First, let’s look at it in general. The characteristic function of the k-th random variable at the point t will be calculated as a sum, where, well, let’s say, l changes from 0 to n_k. First, we have the value, this will be e to the power of itl. See the video. Then, using the Newtonian binomial, we get that this is nothing more than (see the video). So, the characteristic function of the value η will be equal to the product of the characteristic functions of the values ξ_k. It’s given that these values are independent in the population, and their sum is the value of the random variable η. So, the characteristic function of the sum is equal to the product of the characteristic functions, that is, we get the product (see the video).

Here you can see that, for the arbitraries p_k and n_k, this distribution will not be robust, that is, it is not easy for us to convert it to the same form, thus, we will consider the case when all the probabilities are equal. That is, p₁ = p₂ = … = p_m. And q1, q2 ... qm will be equal respectively to 1 minus the corresponding p. Then the index k will be removed, while the n-s will still be left arbitrary. We will get the product of the powers of the same brackets, that is, here these brackets do not depend on k. So, the product of these powers is a given number e to the power of it by p+q, that is the sum of these powers. We get e to the power of it p + q to the power of n₁+ n₂ +, and so on, plus n_m. If this sum is denoted by the number n – this is the total number of tests, we get the characteristic function for the random variable that has the binomial distribution with parameters n and p. It means that if we consider random variables that have the same probability with different parameters n, that is, with different number of tests, such a system of random variables will be robust. They can be combined into one. We study some random event: we conducted a group of tests once, twice, three times. All these tests can easily be combined into one test because the distribution of the sum will also have a binomial distribution where the number of tests is equal to the sum of these particles. This means that the distribution is robust precisely at constant p.

And one more example, for another popular distribution – a continuous distribution, which is called normal. Let me remind you that a random variable has a normal distribution if its distribution density is as follows (see the video). That is, this type has a distribution density for a value that has a normal distribution. To simplify it, we find the distribution density for a random variable that has a standard normal distribution, that is, if m=0, σ² = 1, and σ=1, respectively. Then (I will call it) ѱ0(t) will be equal to (see the video). If we put m=0 and Sigma = 1, we get e to the power of x squared in half and all this in dx. What you can notice is that the exponent is to the power of some second-degree polynomial. We take the full square. And we will take this coefficient out of the sign of the integral (see the video). To have a full square, I need to add (it)².

If we added (it)² here, we need to subtract (it)², see the video. This is dx. Let’s continue, on the next slide (see the video). I will immediately split the part with a square and the part without a square (see the video). Note that e to the power minus t²/2 does not depend on x, it can be taken out of the sign of the integral, which means that only this expression remains under the integral (see the video). If we replace y = x-it, dx will be equal to dy, the boundaries will not change, and, accordingly, this integral will still be equal to √(2π), so we get that the characteristic function of the random variable will be equal to e to the power minus t²/2. This is a characteristic function. But this is for the standard normal distribution.

In general, we use the fact that ѱ_a+bξ(t) is calculated as follows (see the video). We substitute. We know that in a general form our random variable ξ_k will be equal to m _k + σ_k ξ₀, where ξ₀ is the standard normal random variable. Then ѱ_k will be equal to (see the video), and it will be necessary to multiply it by this value, where we will substitute σ_kt instead of t, (see the video). Well. This is what we get. If η is the sum ξ₁ + ξ₂ + … + ξ_m, where the random variables in the population are independent, do we get the same expression? Let’s check it (see the video). Let’s multiply the first multipliers and the second multipliers separately. We get the product of these values (see the video). I can write it down as e to the power of m₁t + m₂t + …, and put t out of the brackets. So I can write that this is the sum m₁+…+m_m multiplied by t.

Further, I will multiply all the second multipliers separately, and again we will get the product (see the video). Since the sum of the random variables, if the first variable has the expected value m₁, and the second one – m₂, and the third one – m_m, the expected value of their sum is equal to the sum of their expected values. Let’s denote this by m_Σ, so as not to confuse it with m. And the sum of the independent variances in the population of random variables will also be equal to the variance of their sums. I will denote it by σ² (without any index). So, I get that the characteristic function for the composition will be as follows: see the video. This means that the characteristic function for the sum will have the same form as the characteristic function for a single random variable. Here it is lost, here it is. That is, it has the same form. This means that the normal distribution is robust for any parameters m and σ. That is, if we have a series of tests with a random value having a normal distribution, another series of tests with normal distribution, and so on, all these series can be combined into one, its expected value is equal to the sum of expected values, the variance is equal to the sum of the variances.

You can start solving the problems where you also need to establish the robustness of standard distributions. Good luck to all of you.