Lecture 2. Special cases of distributions of random variables
This lecture will consider special cases of distributions of random variables.
In practice, we often meet certain types of random variables.
For discrete random variables we have considered such distributions as an alternative distribution, binomial distribution, Poisson distribution.
For continuous random variables there are also most popular distributions, and we have already discussed the first one earlier - it is a uniform distribution.
The random variable is uniformly distributed if all its values lie in thesegment [a, b], and the distribution density for this segment is equal to constant, outside the given segmentthe density will be equal to zero.
Thus the expectation of such a random value is equal to the segment middle, and the variance will be equal to (b – a)2 / 12.
The proof is given in the slide (see video).
Also, quite often in practice we encounter an exponential distribution, for example, it is the basis for the theory of reliability.
A random variable is exponentially distributed, if the values are non-negative, and the distribution density is equal to ae-ax, where a (alpha) is a positive integer, a is a parameter of exponential distribution, that is, for different systems, this parameter will be different.
An important property of exponential distribution is the lack of memory, the same property we have noticed in the geometric distribution when we considered discrete random variables.
I offer for you to prove this property yourself.
Also it is necessary to consider some transformation of random variables, enabling exponential distribution: 1) Let x~E(a), η = ax, , then η ~ E (1); 2) Let x ~ R (0, 1), η = (-1 / a) Inx, then η ~ E (a), as they appear on the slide (see video).
The combination of these two transformations gives us the opportunity to get a random variable having an exponential distribution, when we can generate a uniformly distributed random number.
This is necessary to simulate the operation of some technical devices.
We find the same mean value and variance.
I offer to hold all the calculations on your own.
That is, since the valuesare in the segment from 0 to + ∞, then the integral is precisely within these limits.
Calculate the integrals yourselves, remembering properties of integral calculus.
And the third distribution is called the normal distribution.
It is called so because it is most often found in reality.
Moreover, normal distribution is the limit for binomial, which appears as a result of repeated identical independent experiments.
Therefore, very often, when we consider mathematical statistics, we will addressto a normal distribution and its derivatives.
The random variable has a normal distribution with parameters a andσ2, if σ> 0, if the distribution density has the form shown in the slide (see video).
The function is quite complicated, so it is necessary to remember it.
Moreover, if we take a normal distribution with the parameters 0 and 1(x0~N(0, 1)), we obtain a standard normal distribution.
Its density formula is shown on the slide (cm. Video).
A random variable with normal distribution can take on any real values, while the distribution density has the form shown in the slide (see video).
That is, it has a sufficiently large valuesin some bounded area, and is almost zero outside is.
The more σ2, the flatter is the density.
Let us consider later what it is connected.with.
We introduce Laplace function.
Laplace function is a function that has the form (see video).
Let’s discuss some properties of this function: 1) Ф (0) = 0.5 (at zero, this function is equal to 0.5).
Since the integrand function is not negative, then Ф (x) increases with x, while when x tends to -∞, Ф (x) tends to zero, and if x tends to + ∞, Ф (x) tends to one. 2) Ф (-x) = 1 - F (x).
There is another useful feature that we’ll use, it will also be handy in the future when calculating probabilities.
If we want to find the value of function Ф (x), then we can use MS Excel function NORM.ST.RASP, while it has two options: 1) argument - the point at which we want to calculate the value, 2) zero or one .
If we want to calculate Laplace function, you need to specify parameter one, if we want to calculate differential Laplace function, that is, without integral, then indicate zero.
Note that this function is the integral of the distribution density of the normal random variable, i.e. the distribution function of a standard normal random variable is nothing but Laplace function.
Then the probability of contact with the segment for a given function can be calculated as the difference between Laplace function at points b and a.
Fairly common we encounter a problem of finding the probability of deviation from zero, that is, the probability that a standard distributed random variable deviates from zero by not more than a certain value e.
Using the second property of Laplace function, we find that this probability is calculated as 2Ф (e) - 1, that is, it is important that the values inf the points eand -eare connected.
If we consider that in general, let's introduce a random variable (x-a)/σ, it can be seen that the random variable has a normal distribution with parameters 0 and 1.
Then its distribution function is a Laplace function with a parameter (x- 1)/σ, and the probability of getting this given random variable in the segment ((a; b) is calculated by the formula given in the slide (see the video.).
Similarly, the deviation from the mean value for no more than x, can be calculated as the 2Ф (x / σ) - 1.
Sometimes it is convenient to use this formula directly, sometimes it is convenient to reduce the random variable to a standard random variable.
It is easy to show that for a normal distribution parameter a this is a mean value, and parameter σ2is a variance.