Let the probability density function of random variable X be f (x) = {A / x ^ 2, x > = 10; 0, X

Let the probability density function of random variable X be f (x) = {A / x ^ 2, x > = 10; 0, X

(1) Integral f (x) in interval (- infinity, + infinity) = integral f (x) in interval (10, + infinity)=
=The value of [- A / x] at infinity - the value at x = 10 = A / 10
Let it be equal to zero, that is, let a / 10 = 1, then a = 10
(2) F (x) = the integral of F (x) over the interval (- infinity, x)
When: x = 10: F (x) = the integral of F (x) on the interval (- infinity, x)=
=Integral of 0 on interval (- infinity, 10) + integral of a / x ^ 2 on interval (10, x)
=Integral of 0 + in the interval (10, x] A / x ^ 2=
=[- A / x] value at X - value at x = 10 = A / 10-A / x = 1-10 / X
That is: x = 10, f (x) = 1-10 / X
(3) Let f (k) = 1 / 2. Obviously k > 10,
That is: 1-10 / k = 1 / 2, k = 20

Assuming that the random variables X and y are independent of each other and obey the standard normal distribution, the probability density of the random variable z = x / y is calculated

Joint density function f (x, y) = f (x) * f (y) = (1 / 2 π) e ^ [- (x ^ 2 + y ^ 2) / 2]
Drawing shows that (x is ordinate, y is abscissa)
Yes, Z

Suppose that the random variables X and y are independent of each other and obey n (0,0.5) distribution, then the probability density of Z + y is

It should be the probability density of X + y~
∵ X and y are independent of each other ∵ x + y still obeys normal distribution
∴E(X+Y)=EX+EY=0+0=0
D(X+Y)=DX+DY=0.5+0.5=1
The probability density function of X + y is (let z = x + y): F (z) = (1 / √ 2 π) exp (- Z ^ 2 / 2)

Let the random variable (x, y) be uniformly distributed in the area G surrounded by the curve y = x ^ 2, y = root X. calculate the probability density
Let the random variable (x, y) be uniformly distributed in the area G surrounded by the curve y = x ^ 2, y = root X. (1) write the probability density of (x, y); (2) find the edge probability density FX (x), FY (y); (3) whether X and y are independent of each other?

We only need to get the area of the region g, and the expression of the non-zero part of the probability density of (x, y) is the reciprocal curve of the area of the region g. y = x ^ 2, y = root, X intersection and x = 0, x = 1, and the area is & nbsp; (integral)_ The probability density of 0 ^ 1 (radical x-x ^ 2) DX = 1 / 3, (x, y) is f (x, y) = 3,0 & lt; = x & lt; = 1, X

Given the probability density of random variable (x, y), f (x, y) = a (under r-radical (x ^ 2 + y ^ 2)), x ^ 2 + y ^ 2

f(x,y)=A(R-√(x^2+y^2)),x^2+y^2

Let x ~ n (0,1), find the probability density of y = 2x ^ 2 + 1

The answer is shown below

Let x ~ n (U, σ ^ 2) be a random variable and find the probability density of y = 2x + 5

N(u,σ²),
That is, the density function of X is
fX(x) = 1/(√2π *σ) * e^[-(x-u)² /(2σ²)]
Then y = 2x + 5 ~ n (2U + 5,4 σ ^ 2)
So the probability density of Y is
fY(y)= 1/(√2π *2σ) * e^[-(y-2u-5)² /(8σ²)]

Let x ~ n (0,1), find the probability density of 1: y = e ^ x 2: y = IXI

Calculate probability density with probability function, 1. F = P {y}

Let x ~ U (0,1) be a random variable, and find the probability density of y = x & # 178

P{Y≤y}=P{x^2≤y}=P{-√y≤x≤√y}=1-2P{x≥√y}=1-2(1-P{x≤√y})
=-1+2P{x≤√y}
2F(√y)-1
fY(y)=[F(√y)]'=f(√y)/2√y
F (x) = 1,0 & lt; X & lt; 1; then FYy = 1 / 2 √ y, 0 & lt; Y & lt; 1

The minimum positive period of y = (SiNx) 4 + (cosx) 2 is

y=[(1-cos2x)/2]²+(1+cos2x)/2
=(cos²2x-2cos2x+1)/4+(cos2x+1)/2
=(cos²2x+3)/4
=[(1+cos4x)/2+3]/4
=(cos4x+7)/8
So t = 2 π / 4 = π / 2