Lectures /

# SelfTestSolutions

## SOLUTIONS OF THE SELF TEST PROBLEMS

- Bayes Rule: P(X|E)= P(E|X)P(X)/P(E)

The weatherman has predicted rain tomorrow. In recent years, it has rained only 73 days each year. When it actually rains, the weatherman correctly forecasts rain 70% of the time. When it doesn't rain, he incorrectly forecasts rain 30% of the time. What is the probability that it will rain tomorrow?

The sample space is defined by two mutually-exclusive events - it rains or it does not rain. Additionally, a third event occurs when the weatherman predicts rain. Notation for these events appears below.- Event {$A_{1}$}: It rains tomorrow.
- Event {$A_{2}$}: It does not rain tomorrow.
- Event B: The weatherman predicts rain.

We know the following in terms of probabilities:

- P( A1 ) = 73/365 =0.2 [It rains 73 days out of the year.]
- P( A2 ) = 292/365 = 0.8 [It does not rain 292 days out of the year.]
- P( B | A1 ) = 0.7 [When it rains, the weatherman predicts rain 70% of the time.]
- P( B | A2 ) = 0.3 [When it does not rain, the weatherman predicts rain 30% of the time.]

We want to know P( A1 | B ), the probability it will tomorrow, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below. - P( A1 | B ) = P( A1 ) P( B | A1 ) / [ P( A1 ) P( B | A1 ) + P( A2 ) P( B | A2 ) ]
- P( A1 | B ) = (0.2)(0.7) / [ (0.2)(0.7) + (0.8)(0.3) ]
- P( A1 | B ) = 0.37

- We are given that the pdf is

{$ p(x) = \left\{ \begin{array}{ll} 4x & 0 \le x \le 1/2 \\ -4x+4 & 1/2 \le x \le 1\end{array} \right. $}

What is the equation and graph its cdf*P(x)*?*P(x)*is an anti-derivative of*p(x)*, so integrating the terms of*p(x)*we get:

{$ p(x) = \left\{ \begin{array}{ll} 2x^2 + A & 0 \le x \le 1/2 \\ -2x^2 + 4x + C & 1/2 \le x \le 1\end{array} \right. $}*A*and*C*are the constants of integration.*P(x)*must go to 0 as moved to the left and it is assumed that*P(x)*starts at 0. Therefore*P(0)*= 0 = 2(0)^{2}+*A*; and so*A*= 0.*P(x)*must go to 1 as moved to the right and it is assumed that*P(x)*ends at 1. Therefore*P(1)*= 1 = -2(1)^{2}+ 4(1) +*C*; and so*C*= -1. As a result*P(x)*can be written as:

{$ p(x) = \left\{\begin{array}{ll} 2x^2 & 0 \le x \le 1/2 \\ -2x^2 + 4x - 1 & 1/2 \le x \le 1\end{array} \right.) $}

- Calculate the expected value of {$X$}, {$E[X]$}, where {$X$} is a random variable representing the outcome of a roll of a trick die. Use the sample space {$x \in \{1,2,3,4,5,6\}$} and let

{$p(X=x) = \left[ \begin{array}{ll} 1/2 & x=1 \\ 1/10 & x\neq 1 \end{array} \right] $}.

Recall that the definition of the expected value of a discrete random variable is:

{$E[X]=\sum_{i} p(x_i)x_i $}

Therefore, in this case, we have {$E[X] = 1/2 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 = 25/10 = 2.5$}. Note that 2.5 is**not**a possible outcome, and be aware that expectations may not be in the sample space. - Use the properties of expectation to show that we can rewrite the variance of a random variable
*X*as

{$ Var[X] = E[X^2] - (E[X])^2 $}

If a random variable*X*has the expected value (mean) {$\mu = E[X]$}, then the variance of*X*is given by:

{$ Var[X] =E[(X - \mu)^2] $}

{$ Var[X] = E[X^2 - 2\mu X + \mu^2] $}

{$ Var[X] = E[X^2] - 2\mu E[X] + \mu^2 $}

{$ Var[X] = E[X^2] - 2\mu^2 + \mu^2 $}

{$ Var[X] = E[X^2] - \mu^2 $}

{$ Var[X] = E[X^2] - (E[X])^2 $}

- Consider the following system of equations:

{$ \begin{array}{ccccccc} 2 x_{1} & + & x_{2} & + & x_{3} & = & 3, \\ 4x_{1} & & & + & 2x_{3} & = & 10, \\ 2 x_{1} & + & 2x_{2} & & & = & -2. \end{array} $}

- Write the system as a matrix equation of the form {$Ax = b . $}

{$ A= \left(\begin{array} {ccc} 2 & 1 & 1 \\ 4 & 0 & 2\\ 2 & 2 & 0 \end{array}\right)$} and {$ x= \left(\begin{array} {ccc} x_{1} \\ x_{2}\\ x_{3} \end{array}\right)$} and {$ b= \left(\begin{array} {ccc} 3 \\ 10\\ -2 \end{array}\right)$}

- Write the solution of the system as a column {$S$} and verify by matrix multiplication that {$S $} satisfies the equation {$Ax = b . $}

{$ s= \left(\begin{array} {ccc} 1 \\ -2\\ 3 \end{array}\right)$}

- Write {$b $} as a linear combination of the columns in {$A. $}

{$ b = A_{*1} - 2A_{*2} + 3A_{*3} $}

- Write the system as a matrix equation of the form {$Ax = b . $}
- Consider the following matrix:

{$\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 3\\ 1 & 3 & 4 \end{array}\right]$}- Is the matrix invertible? (Hint: think about singularity and determinants...)

The answer is yes; remember that a matrix is invertible if it is non-singular (i.e. the determinant is not equal to zero). Remember for a 3x3 matrix A:

{$A = \left[\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}]\right] $}

that the determinant is defined as:

{$det(A) = aei + bfg + cdh - afh - ceg - bdi $}

So we have that the determinant of the above matrix is: {$16 + 6 + 9 - 9 - 8 -12 = 2$} - What's the rank of the matrix?

The rank of a matrix is the number of linearly independent columns (or rows, they're interchangeable). For this answer, it's easy, since a matrix cannot be invertible unless it has full rank, meaning all columns(rows) are linearly independent; so {$rk(A) = 3$}. One can calculate rank by performing Gaussian elimination and counting the non-zero rows. See Wikipedia for more information.

- Is the matrix invertible? (Hint: think about singularity and determinants...)
- The eigenvalues of the matrix:

{$ A= \left[\begin{array} {cc} 3 & 6 \\ 1 & 4 \end{array}\right]$} {$\quad$} are {$\lambda = 6$} and {$ \lambda\ = 1.$}

Which of the following is an eigenvector for {$\lambda= 1? $}

Just remember that a vector is an eigenvector for a transformation (i.e. matrix) and an eigenvalue if the following is true:

{$Ax = \lambda x$}

Given that we must find the eigenvector for the eigenvalue of 1, we get:

{$Ax = x$}

We can check each given eigenvector by plugging it into the following system:

{$\begin{array}{ccc}3x + 6y & = & x\\x + 4y & = & y \end{array} $}

This yields {$\left[\begin{array} {c} -3\\ 1\end{array}\right]$} as an eigenvector for {$\lambda = 1.$} - Find the 0, 1, 2 and {$\infty$} norms of

{$ x= \left(\begin{array} {c}2 \\ 1\\-4 \\ -2 \end{array}\right)$} {$\quad$}

{$ \lim_{p \rightarrow 0} ||x||_{p}^{p} $} (The zero norm is the number of non-zero elements.)

{$ ||x||_{1} = \sum_{i=1}^{n} |x_{i}|$} (The grid norm)

{$ ||x||_{2} = \left(\sum_{i=1}^{n} |x_{i}|^{2}\right)^{1/2} $} (The euclidean norm)

{$ ||x||_{\infty} = \lim_{p \rightarrow \infty} ||x||_{p} = \lim_{p \rightarrow \infty} \left(\sum_{i=1}^{n} |x_{i}|^{p}\right)^{1/p} = \max_{i}|x_{i}|$} (The max norm)

The zero norm is 4 since it is the number of non-zero elements.

{$ ||x||_{1} = 9$}

{$ ||x||_{2} = 5$}

{$ ||x||_{\infty} = 4$}