How do you find the probability density function of multiplying two random variables with different

To calculate the probability density function of multiplying two random variables with different distributions, you need to use the convolution formula of the probability density function.

Let the two random variables be x and y, and their probability density functions are fx(x) and fy(y), respectively. Their product z = x * y probability density function fz(z) can be calculated by the following formula:

fz(z) = ∫fx(x) *fy(z / x) *1/x| dx

Where, |1/x|is the reciprocal of the absolute value of x, which means that the calculated probability density function may have different plus and minus signs between different x values.

The core idea of this formula is that for each z-value, we need to consider all the combinations of x and y that can get this z-value, and then multiply and sum their probability density functions.

Note that this formula has a limited scope of application. Specifically, if x and y are independent and identically distributed random variables, then the probability density function of the product z can be calculated using the following formula:

fz(z) = ∫fx(x) *fy(z / x) *1/x| dx

However, if x and y are not independently identically distributed random variables, other, more complex convolution formulas need to be considered.

Knowing the probability density f(x), then it is enough to find f(x) to integrate f(x), and the result of the indefinite integration at x and a is x (b-a), substituting the upper and lower bounds x and a

So the probability of integrating from a to x is (x-a) (b-a) then when x is greater than or equal to b, the probability is equal to 1, so the above equation is obtained.

Probability density functionIt's for continuityRandom variables, suppose that for the continuous random variable x, the distribution function is f(x) and the probability density is f(x).

First of all, for the continuous random variable x, the distribution function f(x) should be continuous, but the function you gave is not continuous at x=-1 and x=1 points, so there is no probability density function, maybe you made a mistake when solving the distribution function.

If f(x) is found correctly, you can calculate the probability density as follows: f(x) = x by definition].

f(y)dy can know f'(x)=f(x), that is, the derivative of the distribution function is equal to the probability density function, so you only need to find the derivative on the basis of the original distribution function to get the probability density function.

Brief introduction. Probability distribution function.

It is one of the basic concepts of probability theory. In practical problems, it is often necessary to study the probability that the value of a random variable is less than a certain value x, and this probability is a function of x, and this function is called the distribution function of random variables, referred to as the distribution function, which is denoted as f(x), that is, f(x)=p (For example, in the design of bridges and dams, the probability that the maximum water level of the river is less than x meters per year is a function of x, and this function is the distribution function of the highest water level. The distribution functions commonly used in practical applications include normal distribution function, Puazon distribution function, binomial distribution function, and so on.

For the distribution function f(x,y) of two-dimensional continuous variables, the definite integral of the probability density function f(x,y) is generally used to solve the problem. For non-continuous variables, they need to be accumulated separately to obtain [similar to the method of finding one-dimensional random variables]. In this problem, when x (0, )y (0, ), the distribution function f(x,y) = (-x)du (-y)f(u,v)dv= (0,x)du (-0,y)2e (-2u-v)dv= (0,x)2e (-2u)du (-0,y)e (-v)dv=[1-e (-2x)][1-e (-y)]. When x (0, )y (0, ), the distribution function f(x,y) = (-0)du (-0)f(u,v)dv=0.

In mathematics, the probability density function of a continuous random variable (which can be shortened to a density function when it is not confusing) is a function that describes the probability of the output value of the random variable near a certain value point.

This refers to one-dimensional continuous random variables, and the same is true for multidimensional continuous variables.

Probability density function of random data: A function that represents the probability that the instantaneous amplitude falls within a specified range, and is therefore the amplitude. It varies with the amplitude of the range taken.

A one-dimensional distribution function and a one-dimensional probability density function for stochastic processes.

A one-dimensional distribution function called an x(t) stochastic process. where p: denotes probability; If it exists:

It is called a one-dimensional probability density function of x(t).

The n-dimensional distribution function and the n-dimensional probability density function of the stochastic process.

Called: an n-dimensional distribution function of x(t).

If it exists: then its x(t) is said to be an n-dimensional probability density.

If for any moment and any n = 1,2 ......Given the distribution function or probability density of x(t), the statistical description of x(t) is considered sufficient.

x and y are independent, calculate the probability of x=x, the probability of y=y, and multiply them directly.

Joint probability distribution, referred to as joint distribution, is the probability distribution of a random variable composed of two or more random variables. According to the number of random ruffle bridgings, the representation of the joint probability distribution is also different. For discrete random variables, the joint probability distribution can be expressed in the form of a list or as a function. For continuous random variables, the joint probability distribution is expressed as an integral of a non-negative function.

Random variable: <> of the space for a given sample

The real-value function on it <>

This is called a (real) random variable. If the value of the random variable x is finite or countless, then x is said to be a discrete random variable. If x is composed of all real numbers or a part of intervals, then x is said to be a continuous random variable, and the values of the continuous random variable are uncountable and infinite.

Random variables are divided into discrete random variables and continuous random variables, and when the probability distribution of random variables is required, they should be treated separately.

1.Discrete joint probability distribution:

For two-dimensional discrete random vectors, let x and y be discrete random variables, <

and <>

If they are all possible geometries of x and y, then the joint probability distribution of x and y can be expressed as a contingency table as shown in the figure on the right, or as a function as shown below<>

Where, |<>

Among the multidimensional random variables, the probability distribution that contains only some of the variables is called the marginal distribution

<>2.Continuous joint probability distribution:

For two-dimensional continuous random vectors, let x and y be continuous random variables, their joint probability distributions, or continuous random variables <>

The probability distribution of <>

<> through a non-negative function

The integral representation of is called the function <>

is the joint probability density. The relationship between the two is as follows:

<> not only completely determines the joint probability distribution of x and y, but also completely determines the probability distribution of x and the probable probability distribution of y, so as to <>

and <>

denotes the probability densities of x and y, respectively, then.

Question 1: The function of a=2,f(x,y) of the normalization of the two-dimensional random distribution is to integrate the density dust function of the two-dimensional random distribution, and the integration region is (frontal and silver brother traces, x) and (y), and the result is **.

Question 2: The method is the same as the first question, and the answer is as follows:

a=1 probability is: 1 3

1. Find the distribution function first

y is definitely distributed over (1,e) and x ln(y) obeys a uniform distribution.

f(x)=p(x<=x)=x；x obeys a uniform distribution on (0,1).

p(ln(y)<=x)=x；Substituting x=ln(y), note that it is a lowercase rush.

p(y<=e^x)=x；The internal condition is transformed to Y as a variable.

p(y<=y)=ln(y)；Substituting x=ln(y), note that it is capitalized.

i.e. f(y) = p(y<=y) = ln(y).

2. Then find the probability density:

f(y)=f'(y)=1/y；The probability density is the derivative of the distribution function.

3. Check the value of the y variable.

There is no overlap and frankness, there is no scattered letter to go beyond it, and the original interpretation is correct.

The two-dimensional distribution of spring dispersion is used to find the mixed second-order partial derivative chop to obtain the density function. As shown below: