TU Wien:Econometrics for Business Informatics VU (Schneider)/Exercise 1 (2024S)

1.1

Let X be a random variable. Give the definition of the expected value or expectation 𝔼(X) when

(a)

X is a discrete random variable taking on values x1, x2, . . . .

Definitions

𝔼(X) = expected value, aka long term average: = µ: = doing an experiment for a long time should be the average

Answer

𝔼(X) = µ = ∑ x * P(x):Sum the products of each value of X and their probability

(b)

X is a continuous random variable with density function f(x) taking on values in ℝ.

Definitions

continuous random variable: a random variable that can take on any value within a certain range or interval

density function: also known as probability density function is a mathematical function that describes the likelihood of a continuous random variable falling within a particular range of values.

• f(x) ≥ 0 for all x in range of X
• $\int \limits _{-\infty }^{\infty }f(x)dx=1$
• P(a ≤ X ≤ b) = $\int \limits _{a}^{b}f(x)dx$

Answer

µ = µ_x = 𝔼(X) = $\int \limits _{-\infty }^{\infty }x*f(x)dx$

1.2

see WS23 Exercise 1.1

1.3

(a)

see WS23 Exercise 1.2

(b)

Let X be uniformly distributed on [0,1], that is, X has density function

$f(x)={\begin{cases}1,&{\text{x ∈ [0,1]}}\\0,&{\text{x ∉ [0,1].}}\end{cases}}$

(We write X ~ unif[0,1].) Calculate 𝔼(X) and Var(X).

Uniformly distributed therefore this formula:

𝔼(X) = $\int \limits _{-\infty }^{\infty }x*f(x)dx$

=$\int \limits _{-\infty }^{0}x*0dx+\int \limits _{0}^{1}x*1dx+\int \limits _{1}^{\infty }x*0dx$

= $0+\int \limits _{0}^{1}xdx+0$

= $[{\frac {x^{2}}{2}}]$₀¹

= ${\frac {1}{2}}-0$

𝔼(X) = ${\frac {1}{2}}$

Var(X) = 𝔼(X²) - 𝔼(X)²

Calculation for 𝔼(X²)

𝔼(X²) = $\int \limits _{-\infty }^{\infty }x^{2}*f(x)dx$

= $\int \limits _{0}^{1}x^{2}*1dx$

= $[{\frac {x^{3}}{3}}]$₀¹

= ${\frac {1}{3}}-0$

= ${\frac {1}{3}}$

Var(X) = ${\frac {1}{3}}-({\frac {1}{2}})^{2}$

= ${\frac {1}{3}}-{\frac {1}{4}}={\frac {1}{12}}$

(c)

For X ~ N(µ, σ2), specify the density as well as 𝔼(X) and Var(X). (You do not need to calculate expectation and variance.)

N(µ, σ²) = normally distributed random variable X with mean µ and variance σ²

density function(PDF, given by the normal distribution formula):$f(x)={\frac {1}{\sqrt {2\pi \sigma }}}e^{\frac {-(x-\mu )^{2}}{2\sigma ^{2}}}$

𝔼(X) = µ

Var(X) = σ²

1.4

Let X1, ..., Xn be independent and identically distributed random variables with 𝔼(Xi) = µ.

(a)

What is the "difference" between µ and ${\bar {X}}={\frac {1}{n}}\sum \limits _{i=1}^{n}X$i?

The difference between µ and ${\bar {X}}$ lies in their interpretation and their roles in statistics:

• µ is a population parameter, representing the true average value of the entire population
• ${\bar {X}}$ is a sample statistic, providing an estimate of µ based on sample of observations.

(b)

Show that ${\bar {X}}={\frac {1}{n}}\sum \limits _{i=1}^{n}X$_i is an unbiased estimator for µ.

unbiased estimator = a statistical value that provides an estimate of a population parameter without systematically under or over estimating an average.

We know µ = average value

=> if 𝔼(X) = µ then ${\bar {X}}$ is an unbiased estimator

𝔼(${\bar {X}}$) = 𝔼(${\frac {1}{n}}\sum \limits _{i=1}^{n}X$_i)

since X₁, ..., X_n are independent and identically distributed, we can use the linearity of expectations

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}\mathbb {E} (X$_i)

since all X_i are identically distributed, their mean is equal to µ

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}\mu$

= ${\frac {1}{n}}*n*\mu$

= µ

Therefore ${\bar {X}}={\frac {1}{n}}\sum \limits _{i=1}^{n}X$_i is an unbiased estimator for µ, as its expected value equals µ.

1.5

The sample covariance Sxy of numbers x1, ..., xn and y1, ..., yn is defined by

$S_{\text{xy}}={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})$

Show that $S_{\text{xy}}={\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*(y_{i}-{\bar {y}})={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})*y_{i}={\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\bar {x}}*{\bar {y}}$.

$S_{\text{xy}}={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})={\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*(y_{i}-{\bar {y}})$

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}*y_{i}-x_{i}*{\bar {y}}-{\bar {x}}*y_{i}+{\bar {x}}*{\bar {y}})$

Using linearity of sums

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}}-{\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*y_{i}+{\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*{\bar {y}}$

Calculation1 for ${\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*y_{i}$

Because - ${\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}={\bar {x}}$

= ${\bar {x}}*({\frac {1}{n}}\sum \limits _{i=1}^{n}y_{i})$

Because - ${\frac {1}{n}}\sum \limits _{i=1}^{n}y_{i}={\bar {y}}$

= ${\bar {x}}{\bar {y}}$

Calculation2 for ${\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*{\bar {y}}$

see Calculation1 step 1

= ${\bar {x}}{\bar {y}}$

= $({\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i})-({\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}})-{\bar {x}}*{\bar {y}}+{\bar {x}}*{\bar {y}}$

=${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}}$

Using linearity of sums

=${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-x_{i}*{\bar {y}}$

Using factoring out the common factor

=${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*(y_{i}-{\bar {y}})$

q.e.d. Formula 1

$S_{\text{xy}}={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})={\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\bar {x}}*{\bar {y}}$

proof as in 1 until

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}}$

Calculation3 for ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}}$

see Calculation2

= ${\bar {x}}*{\bar {y}}$

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\bar {x}}{\bar {y}}$

q.e.d. Formula 3

$S_{\text{xy}}={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})(y_{i}-{\bar {y}})={\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})*y_{i}$

proof as in 1 until

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*{\bar {y}}-{\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*y_{i}+{\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*{\bar {y}}$

using calculation2 and calculation3

= $({\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i})-{\bar {x}}*{\bar {y}}-({\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*y_{i})+{\bar {x}}*{\bar {y}}$

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\frac {1}{n}}\sum \limits _{i=1}^{n}{\bar {x}}*y_{i}$

using linearity of sums

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}x_{i}*y_{i}-{\bar {x}}*y_{i})$

using factoring our the common factor

= ${\frac {1}{n}}\sum \limits _{i=1}^{n}(x_{i}-{\bar {x}})*y_{i}$

q.e.d. Formula 2

1.6

(a)

Find concrete matrices A and B such that AB ≠ BA. With these matrices, show that (AB)' = B'A'.

AB ≠ BA:

A = ${\begin{pmatrix}1&2\\3&4\end{pmatrix}}$, B = ${\begin{pmatrix}5&6\\7&8\end{pmatrix}}$

A*B = ${\begin{pmatrix}1*5+2*7&1*6+2*8\\3*5+4*7&3*6+4*8\end{pmatrix}}={\begin{pmatrix}19&22\\43&50\end{pmatrix}}$

B*A = ${\begin{pmatrix}5*1+6*3&5*2+6*4\\7*1+8*3&7*2+8*4\end{pmatrix}}={\begin{pmatrix}23&34\\31&46\end{pmatrix}}$

AB ≠ BA: ${\begin{pmatrix}19&22\\43&50\end{pmatrix}}$ ≠ ${\begin{pmatrix}23&34\\31&46\end{pmatrix}}$

(AB)' = B'A':

(AB)' = transpose of Matrix

=${\begin{pmatrix}19&22\\43&50\end{pmatrix}}'$= ${\begin{pmatrix}19&43\\22&50\end{pmatrix}}$

A' = ${\begin{pmatrix}1&2\\3&4\end{pmatrix}}$' = ${\begin{pmatrix}1&3\\2&4\end{pmatrix}}$

B' = ${\begin{pmatrix}5&6\\7&8\end{pmatrix}}$' = ${\begin{pmatrix}5&7\\6&8\end{pmatrix}}$

B'A' = ${\begin{pmatrix}5&7\\6&8\end{pmatrix}}$ * ${\begin{pmatrix}1&3\\2&4\end{pmatrix}}$

= ${\begin{pmatrix}5*1+7*2&5*3+7*4\\6*1+8*2&6*3+8*4\end{pmatrix}}$

= ${\begin{pmatrix}19&43\\22&50\end{pmatrix}}$

(AB)' = B'A'

${\begin{pmatrix}19&43\\22&50\end{pmatrix}}$ = ${\begin{pmatrix}19&43\\22&50\end{pmatrix}}$

(b)

Let

 A = ${\begin{pmatrix}1&0\\2&1\end{pmatrix}}$

and compute A', A-1 , (A')-1, (A-1)'.

A' = transpose of a matrix

= ${\begin{pmatrix}1&2\\0&1\end{pmatrix}}$

A^-1 = inverse of a matrix

→ with Elementary Transformation Method

= ${\Biggl (}{\begin{matrix}1&0\\2&1\end{matrix}}{\Biggl |}{\begin{matrix}1&0\\0&1\end{matrix}}{\Biggr )}$ / - (2 * first line)

= ${\Biggl (}{\begin{matrix}1&0\\0&1\end{matrix}}{\Biggl |}{\begin{matrix}1&0\\-2&1\end{matrix}}{\Biggr )}$

A^-1 = ${\begin{pmatrix}1&0\\-2&1\end{pmatrix}}$

(A')^-1 = ${\Biggl (}{\begin{matrix}1&2\\0&1\end{matrix}}{\Biggl |}{\begin{matrix}1&0\\0&1\end{matrix}}{\Biggr )}$ / -(2 * second line)

= ${\Biggl (}{\begin{matrix}1&0\\0&1\end{matrix}}{\Biggl |}{\begin{matrix}1&-2\\0&1\end{matrix}}{\Biggr )}$

(A')^-1 = ${\begin{pmatrix}1&-2\\0&1\end{pmatrix}}$

(A^-1)' = ${\begin{pmatrix}1&0\\-2&1\end{pmatrix}}'$

= ${\begin{pmatrix}1&-2\\0&1\end{pmatrix}}$

(c)

Let 

 A = ${\begin{pmatrix}1&0&5\\0&23&5\\1&24&0\\0&0&0\end{pmatrix}}$, b = ${\begin{pmatrix}3\\0\\1\end{pmatrix}}$

and compute Ab. Demonstrate that the result is a linear combination of the columns of A with the coefficients being the components of b.