Kuzee
Probability and Mathematical Statistics

Comprehensive Exercises on the Numerical Characteristics of Random Variables

Exercises

Exercise 1

Given the distribution law P(X=1)=0.2,P(X=2)=0.5,P(X=3)=0.3P(X=1)=0.2, P(X=2)=0.5, P(X=3)=0.3 of XX, find E(X)E(X).

mathematical expectationdiscrete random variabledefinition and computation

Approach: Use the definition of the mathematical expectation of a discrete random variable.

Steps:

  1. E(X)=∑i=1∞xipiE(X) = \sum_{i=1}^{\infty} x_i p_i
  2. =1×0.2+2×0.5+3×0.3= 1 \times 0.2 + 2 \times 0.5 + 3 \times 0.3
  3. =0.2+1.0+0.9=2.1= 0.2 + 1.0 + 0.9 = 2.1

Answer: 2.12.1

Exercise 2

Given E(X)=2,D(X)=3E(X)=2, D(X)=3, find E(3X−1)E(3X-1) and D(3X−1)D(3X-1).

mathematical expectationvarianceapplication of properties

Approach: Use the properties of mathematical expectation and the variance.

Steps:

  1. E(3X−1)=3E(X)−1=3×2−1=5E(3X-1) = 3E(X) - 1 = 3 \times 2 - 1 = 5
  2. D(3X−1)=32D(X)=9×3=27D(3X-1) = 3^2 D(X) = 9 \times 3 = 27

Answer: E(3X−1)=5E(3X-1) = 5, D(3X−1)=27D(3X-1) = 27

Exercise 3

Given that X,YX, Y are independent with E(X)=1,E(Y)=2E(X)=1, E(Y)=2, find E(X+Y)E(X+Y).

mathematical expectationlinearity of expectationcomputation

Approach: Use the linearity of mathematical expectation.

Steps:

  1. E(X+Y)=E(X)+E(Y)E(X+Y) = E(X) + E(Y)
  2. =1+2=3= 1 + 2 = 3
  3. The linearity of expectation holds for any random variables, so independence is not required as a premise in this step.

Answer: E(X+Y)=3E(X + Y) = 3.

Exercise 4

Given E(X)=1,E(Y)=2,Cov(X,Y)=3,σX=2,σY=1E(X)=1, E(Y)=2, Cov(X,Y)=3, \sigma_X=2, \sigma_Y=1, find ρXY\rho_{XY}.

correlation coefficientdefinition and computationstandardization

Approach: Use the definition of the correlation coefficient.

Steps:

  1. ρXY=Cov(X,Y)σXσY\rho_{XY} = \frac{Cov(X,Y)}{\sigma_X \sigma_Y}
  2. =32×1=1.5= \frac{3}{2 \times 1} = 1.5

Answer: 1.51.5

Exercise 5

Given the distribution function F(x)={0,x<0x,0≤x<11,x≥1F(x)=\begin{cases} 0, x<0 \\ x, 0\leq x<1 \\ 1, x\geq1 \end{cases} of XX, find E(X)E(X).

mathematical expectationdistribution functioncontinuous random variable

Approach: First find the density function, then compute the expectation.

Steps:

  1. f(x)=F′(x)={1,0≤x<10,otherwisef(x) = F'(x) = \begin{cases} 1, & 0 \leq x < 1 \\ 0, & \text{otherwise} \end{cases}
  2. E(X)=∫01x⋅1dx=x22∣01=12E(X) = \int_0^1 x \cdot 1 dx = \frac{x^2}{2}\big|_0^1 = \frac{1}{2}

Answer: 12\frac{1}{2}

Exercise 6

Given that the random variable XX follows an exponential distribution with parameter λ\lambda, find E(X)E(X) and D(X)D(X).

mathematical expectationexponential distributionexpectation and variance

Approach: Use the mathematical expectation and variance formulas of the exponential distribution.

Steps:

  1. The mathematical expectation of the exponential distribution: E(X)=1λE(X) = \frac{1}{\lambda}
  2. The variance of the exponential distribution: D(X)=1λ2D(X) = \frac{1}{\lambda^2}

Answer: E(X)=1λE(X) = \frac{1}{\lambda}, D(X)=1λ2D(X) = \frac{1}{\lambda^2}

Exercise 7

Given that the random variable XX follows a binomial distribution with parameters n,pn, p, find E(X)E(X) and D(X)D(X).

mathematical expectationbinomial distributionexpectation and variance

Approach: Use the mathematical expectation and variance formulas of the binomial distribution.

Steps:

  1. The mathematical expectation of the binomial distribution: E(X)=npE(X) = np
  2. The variance of the binomial distribution: D(X)=np(1−p)D(X) = np(1-p)

Answer: E(X)=npE(X) = np, D(X)=np(1−p)D(X) = np(1-p)

Exercise 8

Given that the joint distribution law of the random variables XX and YY is:

X\YX \backslash Y01
00.30.2
10.20.3

Find Cov(X,Y)Cov(X,Y).

covariancecomputation formulajoint distribution law

Approach: Use the simplified computation formula for the covariance.

Steps:

  1. E(X)=0×0.5+1×0.5=0.5E(X) = 0 \times 0.5 + 1 \times 0.5 = 0.5
  2. E(Y)=0×0.5+1×0.5=0.5E(Y) = 0 \times 0.5 + 1 \times 0.5 = 0.5
  3. E(XY)=0×0×0.3+0×1×0.2+1×0×0.2+1×1×0.3=0.3E(XY) = 0 \times 0 \times 0.3 + 0 \times 1 \times 0.2 + 1 \times 0 \times 0.2 + 1 \times 1 \times 0.3 = 0.3
  4. Cov(X,Y)=E(XY)−E(X)E(Y)=0.3−0.5×0.5=0.05Cov(X,Y) = E(XY) - E(X)E(Y) = 0.3 - 0.5 \times 0.5 = 0.05

Answer: 0.050.05

Exercise 9

Given that XX and YY are independent, D(X)=4D(X) = 4, and D(Y)=9D(Y) = 9, find D(X+Y)D(X + Y).

varianceindependence and additivityvariance computation

Approach: Use the additivity of the variance for independent random variables.

Steps:

  1. If XX and YY are independent, then D(X+Y)=D(X)+D(Y)D(X + Y) = D(X) + D(Y)
  2. =4+9=13= 4 + 9 = 13
  3. Independence is exactly the premise of the additivity of the variance; here XX and YY are independent, so the two variances can be added directly.

Answer: D(X+Y)=13D(X + Y) = 13.

Exercise 10

Given that the random variable XX follows a Poisson distribution with parameter λ\lambda, find E(X)E(X) and D(X)D(X).

mathematical expectationPoisson distributionexpectation and variance

Approach: Use the mathematical expectation and variance formulas of the Poisson distribution.

Steps:

  1. The mathematical expectation of the Poisson distribution: E(X)=λE(X) = \lambda
  2. The variance of the Poisson distribution: D(X)=λD(X) = \lambda

Answer: E(X)=λE(X) = \lambda, D(X)=λD(X) = \lambda

Exercise 11

Given that the random variable XX follows a normal distribution with parameters μ,σ2\mu, \sigma^2, find E(X)E(X) and D(X)D(X).

mathematical expectationnormal distributionexpectation and variance

Approach: Use the mathematical expectation and variance formulas of the normal distribution.

Steps:

  1. The mathematical expectation of the normal distribution: E(X)=μE(X) = \mu
  2. The variance of the normal distribution: D(X)=σ2D(X) = \sigma^2

Answer: E(X)=μE(X) = \mu, D(X)=σ2D(X) = \sigma^2

Exercise 12

Given that the correlation coefficient of the random variables XX and YY is 0.80.8, D(X)=4D(X) = 4, and D(Y)=9D(Y) = 9, find Cov(X,Y)Cov(X,Y).

correlation coefficientdefinition and computationstandardization

Approach: Use the definition of the correlation coefficient.

Steps:

  1. ρXY=Cov(X,Y)D(X)D(Y)\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{D(X)D(Y)}}
  2. 0.8=Cov(X,Y)4×9=Cov(X,Y)60.8 = \frac{Cov(X,Y)}{\sqrt{4 \times 9}} = \frac{Cov(X,Y)}{6}
  3. Cov(X,Y)=0.8×6=4.8Cov(X,Y) = 0.8 \times 6 = 4.8

Answer: 4.84.8

Exercise 13

Given that the random variable XX follows the uniform distribution on the interval [a,b][a, b], find E(X)E(X) and D(X)D(X).

mathematical expectationuniform distributionexpectation and variance

Approach: Use the mathematical expectation and variance formulas of the uniform distribution.

Steps:

  1. The mathematical expectation of the uniform distribution: E(X)=a+b2E(X) = \frac{a+b}{2}
  2. The variance of the uniform distribution: D(X)=(b−a)212D(X) = \frac{(b-a)^2}{12}

Answer: E(X)=a+b2E(X) = \frac{a+b}{2}, D(X)=(b−a)212D(X) = \frac{(b-a)^2}{12}

Exercise 14

Given the distribution law P(X=1)=0.2,P(X=2)=0.5,P(X=3)=0.3P(X=1)=0.2, P(X=2)=0.5, P(X=3)=0.3 of XX, find D(X)D(X).

variancecovariance formulacomputation

Approach: Use the simplified computation formula for the variance.

Steps:

  1. E(X)=1×0.2+2×0.5+3×0.3=2.1E(X) = 1 \times 0.2 + 2 \times 0.5 + 3 \times 0.3 = 2.1
  2. E(X2)=12×0.2+22×0.5+32×0.3=0.2+2.0+2.7=4.9E(X^2) = 1^2 \times 0.2 + 2^2 \times 0.5 + 3^2 \times 0.3 = 0.2 + 2.0 + 2.7 = 4.9
  3. D(X)=E(X2)−[E(X)]2=4.9−(2.1)2=4.9−4.41=0.49D(X) = E(X^2) - [E(X)]^2 = 4.9 - (2.1)^2 = 4.9 - 4.41 = 0.49

Answer: 0.490.49

Exercise 15

Prove that if XX and YY are independent, then ρXY=0\rho_{XY} = 0.

correlation coefficientindependenceproof

Approach: Use the fact that the covariance of independent random variables is zero.

Steps:

  1. If XX and YY are independent, then Cov(X,Y)=0Cov(X,Y) = 0
  2. Hence ρXY=Cov(X,Y)D(X)D(Y)=0D(X)D(Y)=0\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{D(X)D(Y)}} = \frac{0}{\sqrt{D(X)D(Y)}} = 0

Answer: the proof is complete

Previous
Covariance and the Correlation Coefficient
Next
Law of Large Numbers and Central Limit Theorem

Found an issue with this content?

Tell us what could be improved.