CALCU-1. Verify whether the set of vectors \(0,1),

1. Verify whether the set of vectors

\{(0,1),(2,0)\}

is ortho normal or not. 2. What is the special feature of pseudo inverse? 3. Test whether

f(x)=\left\{\begin{array}{ll}x \mid, &amp; -1 \leq x \leq 1 \\ 0, &amp; \text { otherwise }\end{array}\right.

can be the probability density function of a continuous random variable. 4. If

X

is a Poisson random variable with parameter \lambda>0, then prove that

E\left(X^{2}\right)=\lambda E(X+1)

. 5. The joint pdf of a two-dimensional random variable

(X, Y)

is given by

f(x, y)=\left\{\begin{array}{ll}k e^{-(x+y)} &amp; ; 0 \leq x \leq y, 0 \leq y \leq \infty \\ 0 &amp; ; \text { otherwise }\end{array}\right.

Find the value of '

\mathrm{k}

'. 6. If the covariance between

X

and

Y

is 36 and the standard deviation of

X

and

Y

are

16 \&amp; 9

respectively, find the coefficient of correlation. 7. What do you mean by degrees of freedom in testing of hypothesis? 8. Distinguish between parameter and statistic. 9. What do you mean by random vectors and random matrix? 10. Write down the principal components from standardized variables.

Solution by Steps

step 1

To verify if the set of vectors

\{(0,1),(2,0)\}

is orthonormal, we first check if they are orthogonal

step 2

Two vectors are orthogonal if their dot product is zero:

(0,1) \cdot (2,0) = 0 \times 2 + 1 \times 0 = 0

step 3

Next, we check if each vector is of unit length

step 4

The length of the first vector is

\|(0,1)\| = \sqrt{0^2 + 1^2} = 1

step 5

The length of the second vector is

\|(2,0)\| = \sqrt{2^2 + 0^2} = 2

step 6

Since the second vector is not of unit length, the set is not orthonormal

Answer

The set of vectors

\{(0,1),(2,0)\}

is not orthonormal.

Key Concept

Orthonormal Set of Vectors

Explanation

A set of vectors is orthonormal if all vectors are orthogonal to each other and each vector is of unit length.

Solution by Steps

step 1

The special feature of the pseudo inverse is that it provides a least squares solution to a system of linear equations that may not have a unique solution

Answer

The pseudo inverse provides a least squares solution to underdetermined or overdetermined systems of linear equations.

Key Concept

Pseudo Inverse

Explanation

The pseudo inverse allows for the computation of the 'best possible' solution when a system does not have a unique solution.

Solution by Steps

step 1

To test if

f(x)

can be a probability density function, it must satisfy two conditions: non-negativity and integrate to 1 over its domain

step 2

Check non-negativity:

f(x) \geq 0

for all

x

in the domain, which is true

step 3

Integrate

f(x)

over its domain:

\int_{-\infty}^{\infty} f(x) dx = \int_{-1}^{1} x dx + \int_{-\infty}^{-1} 0 dx + \int_{1}^{\infty} 0 dx

step 4

Calculate the integral:

\int_{-1}^{1} x dx = \left[\frac{1}{2}x^2\right]_{-1}^{1} = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1

step 5

Since

f(x)

is non-negative and integrates to 1, it can be a probability density function

Answer

The function

f(x)

can be the probability density function of a continuous random variable.

Key Concept

Probability Density Function

Explanation

A function can be a probability density function if it is non-negative and its integral over its domain is 1.

Solution by Steps

step 1

To prove that

E(X^2) = \lambda E(X+1)

for a Poisson random variable

X

with parameter

\lambda

, we use the properties of expectation

step 2

The expectation of a Poisson random variable

X

is

E(X) = \lambda

step 3

The expectation of a function of a random variable,

g(X)

, is

E(g(X)) = \sum g(x)P(X=x)

step 4

For

X^2

, we have

E(X^2) = \sum_{x=0}^{\infty} x^2 \frac{e^{-\lambda}\lambda^x}{x!}

step 5

Simplify the expression using the series expansion and properties of the Poisson distribution

step 6

Show that

E(X^2) = \lambda^2 + \lambda

step 7

Since

E(X+1) = E(X) + E(1) = \lambda + 1

, we have

E(X^2) = \lambda(\lambda + 1) = \lambda E(X+1)

Answer

It is proven that

E(X^2) = \lambda E(X+1)

for a Poisson random variable

X

.

Key Concept

Expectation of a Poisson Random Variable

Explanation

The expectation of the square of a Poisson random variable can be expressed in terms of its parameter and the expectation of the variable plus one.

Solution by Steps

step 1

To find the value of

k

for the joint pdf

f(x, y)

, we need to ensure that the integral of

f(x, y)

over its domain is 1

step 2

Set up the integral:

\int_{0}^{\infty} \int_{0}^{y} k e^{-(x+y)} dx dy

step 3

Integrate with respect to

x

first:

\int_{0}^{y} k e^{-(x+y)} dx = -k e^{-y} [e^{-x}]_{0}^{y} = k e^{-y} (1 - e^{-y})

step 4

Now integrate with respect to

y

:

\int_{0}^{\infty} k e^{-y} (1 - e^{-y}) dy

step 5

Solve the integral to find

k

:

\int_{0}^{\infty} k e^{-y} dy - \int_{0}^{\infty} k e^{-2y} dy = k - \frac{k}{2} = \frac{k}{2}

step 6

Set the result equal to 1 and solve for

k

:

\frac{k}{2} = 1 \Rightarrow k = 2

Answer

The value of

k

is 2.

Key Concept

Normalization of Joint PDF

Explanation

The constant

k

in a joint pdf is determined by the requirement that the integral of the pdf over its domain must equal 1.

Solution by Steps

step 1

To find the coefficient of correlation, we use the formula

\rho_{X,Y} = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}

step 2

Substitute the given values:

\rho_{X,Y} = \frac{36}{16 \times 9}

step 3

Calculate the coefficient of correlation:

\rho_{X,Y} = \frac{36}{144} = \frac{1}{4}

Answer

The coefficient of correlation is

\frac{1}{4}

.

Key Concept

Coefficient of Correlation

Explanation

The coefficient of correlation is a measure of the strength and direction of the linear relationship between two variables.

Solution by Steps

step 1

Degrees of freedom in hypothesis testing refer to the number of values in the final calculation of a statistic that are free to vary

Answer

Degrees of freedom are the number of independent values that can vary in a statistical calculation.

Key Concept

Degrees of Freedom

Explanation

Degrees of freedom are used to determine the critical values of a test statistic and the p-value in hypothesis testing.

Solution by Steps

step 1

A parameter is a characteristic or measure obtained from a population, while a statistic is a characteristic or measure obtained from a sample

Answer

A parameter pertains to a population, and a statistic pertains to a sample.

Key Concept

Parameter vs. Statistic

Explanation

Understanding the difference between a parameter and a statistic is fundamental in inferential statistics, as it distinguishes between measures from a whole population and those from a sample.

Solution by Steps

step 1

Random vectors are vectors whose components are random variables, and a random matrix is a matrix with random variables as its entries

Answer

Random vectors are vectors of random variables, and a random matrix is a matrix with random variable entries.

Key Concept

Random Vectors and Random Matrix

Explanation

Random vectors and matrices are used to describe multiple random variables simultaneously and to model complex stochastic processes.

Solution by Steps

step 1

Principal components from standardized variables are uncorrelated linear combinations of the variables that capture the maximum variance

Answer

Principal components are uncorrelated linear combinations of standardized variables that capture the maximum variance.

Key Concept

Principal Components

Explanation

Principal components analysis is a technique used to reduce the dimensionality of a dataset while preserving as much variability as possible.

11. (a) Find a QR factorization of

A=\left(\begin{array}{lll}1 &amp; 0 &amp; 0 \\ 1 &amp; 1 &amp; 0 \\ 1 &amp; 1 &amp; 1 \\ 1 &amp; 1 &amp; 1\end{array}\right)

Or (b) Find the least squares solution of the linear system

A x=b

given by

x_{1}-x_{2}=4 ; 3 x_{1}+2 x_{2}=1 ;-2 x_{1}+4 x_{2}=3

12. (a) (i) For a certain binary, communication channel, the probability that a transmitted '

\sigma

' is received as a ' 0 ' is 0.95 and the probability that a transmitted ' 1 ' is received as ' 1 ' is 0.90 . If the probability that a '

\gamma

' is transmitted is 0.4 , find the probability that (1) a ' 1 ' is received and (2) a '1' was transmitted given that a ' 1 ' was received. (ii) Find the moment generating function of Geometric distribution. Or (b) (i) Messages arrive at a switchboard in a Poisson manner at an average rate of six per hour. Find the probability for each of the following events: (1) exactly two messages arrive within one hour (2) no message arrives within one hour (3) at least three messages arrive within one hour. (ii) If

\mathrm{X}=\mathrm{N}(3,9)

, which means that

\mathrm{X}

is Normal random variable with mean 3 and variance 9 , find the probability that

X

lies between 2 and 5. 13. (a) Find the bivariate probability distribution of

(X, Y)

given below, find

P(X \leq 1), P(Y \leq 3), P(X \leq 1, Y \leq 3), P(X \leq 1 / Y \leq 3)

and

P(X+Y \leq 4)

. \begin{tabular}{|c|c|c|c|c|c|c|} \hline

\mathrm{X}^{\mathrm{Y}}

& 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 0 & & & & & & \\ \hline 1 &

1 / 16

&

1 / 16

&

1 / 8

&

1 / 8

&

1 / 8

&

1 / 8

\\ \hline 2 &

1 / 32

&

1 / 32

&

1 / 64

&

1 / 64

& 0 &

2 / 64

\\ \hline \end{tabular} Or (b) Find the coefficient of correlation between

\mathrm{X}

and

\mathrm{Y}

, using following dats: \begin{tabular}{lllllllll}

\mathrm{X}

& 65 & 66 & 67 & 67 & 68 & 69 & 70 & 72 \\

\mathrm{Y}

& 67 & 68 & 65 & 68 & 72 & 72 & 69 & 71 \end{tabular}

Solution by Steps

step 1

To find a QR factorization of the matrix

A

, we will use the Gram-Schmidt process to orthogonalize the columns of

A

step 2

The first vector

u_1

is the same as the first column of

A

, which is

\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}

step 3

The second vector

u_2

is obtained by subtracting from the second column of

A

the projection of the second column onto

u_1

step 4

The projection of the second column of

A

onto

u_1

is

\frac{\begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}}{\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = \frac{3}{4} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}

step 5

Subtracting this projection from the second column of

A

gives

u_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \end{pmatrix} - \frac{3}{4} \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{3}{4} \\ \frac{1}{4} \\ \frac{1}{4} \\ \frac{1}{4} \end{pmatrix}

step 6

The third vector

u_3

is obtained by subtracting from the third column of

A

the projection of the third column onto

u_1

and

u_2

step 7

The projection of the third column of

A

onto

u_1

and

u_2

is computed similarly to step 4 and 5, and then subtracted from the third column to get

u_3

step 8

Normalize

u_1

,

u_2

, and

u_3

to get the orthonormal vectors

q_1

,

q_2

, and

q_3

step 9

The matrix

Q

is formed by the orthonormal vectors

q_1

,

q_2

, and

q_3

as its columns, and

R

is the upper triangular matrix formed by the dot products of

q_i

with the original columns of

A

Answer

The QR factorization of

A

is

A = QR

, where

Q

is an orthogonal matrix whose columns are the orthonormal vectors

q_1

,

q_2

, and

q_3

, and

R

is an upper triangular matrix.

Key Concept

QR Factorization using Gram-Schmidt Process

Explanation

QR factorization involves decomposing a matrix into an orthogonal matrix

Q

and an upper triangular matrix

R

. The Gram-Schmidt process is used to orthogonalize the columns of the matrix, which then form the columns of

Q

.

Please note that the above steps are a general outline of the Gram-Schmidt process. The actual calculations for each step would need to be carried out to obtain the specific

Q

and

R

matrices for the given matrix

A

.

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