Probabilistic Methods for Solving Density Functions

Probability theory. Multivariate random variable density function solution

(1) It is known that f(x)=1, (0=0), Z is greater than 0

Then F(z)=P(X Y

Draw the integral interval on the coordinate axis

That is, when 0

When z>=1, the x integral interval is (0,1), and the y integral interval is (0,z-x)

Integrate f(x)*f(y)=e^(-y) in the above interval, we have

0

When z>=1, F(z)=e^(-z)-e^(1-z) 1

Guide, yes

When 0

When z>=1, f(z)=e^(1-z)-e^(-z)

Therefore, the probability density function of Z is

f(z)=0,z

f(z)=1-e^(-z),0

f(z)=e^(1-z)-e^(-z), when z>=1

(2)F(z))=P(-2lnXe^(-z/2))

When z

When z>=0, integrate f(x) from e^(-z/2) to 1, and get F(z)=1-e^(-z/2)

Guide, yes

f(z)=e^(-z/2)/2

Therefore, the probability density function of Z is

f(z)=0,z

f(z)=e^(-z/2)/2,z>=0

Please help experts to solve density function exercises

1. Because the double integral of the joint density function is 1, it is uniformly distributed on the circle, so

f(x,y)= 1/(pi*R*R) ,x^2 y^2=0, other areas

2. The edge density function of x is defined by f Equivalent to integrating a constant, the integration interval is related to x, y1, y2 are the ordinates of the point on the circle with the abscissa of x)

=1/(pi*R*R) * 2 * Root sign (R^2-x^2)

For the edge density function of y, just replace x in the formula with y

3 Under the condition {X= x}, the conditional density function is defined as f Y|X(y|x) =f(x,y)/f(x) =1/2* root sign (R ^2-x^2) (substitute the conclusions of the previous two questions)

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