Exercise 6 If X Is A Continuous Random Variable With A Probability Density Function F(x) = C.sina: 0 (2024)

(a) An orthonormal basis for R^3 using the Gram-Schmidt orthogonalization procedure for set B1 is: ((1/√2, 0, 1/√2), (1/√6, 2/√6, 1/√6), (-1/√3, 2/√3, -1/√3)).

(b) An orthonormal basis for R^3 using the Gram-Schmidt orthogonalization procedure for set B2 is: ((2/√6, 1/√6, 1/√6), (1/√6, -1/√6, √2/√6), (-1/√17, 1/√17, 2/√17)).

(a) Applying the Gram-Schmidt orthogonalization procedure to set B1 = {(1,0,1),(1,1,0),(1,1,2)}:

Step 1: Normalize the first vector:

v1 = (1,0,1)

u1 = v1 / ||v1|| = (1,0,1) / √(1^2 + 0^2 + 1^2) = (1,0,1) / √2 = (√2/2, 0, √2/2)

Step 2: Compute the projection of the second vector onto the subspace spanned by u1:

v2 = (1,1,0)

proj = (v2 · u1) / (u1 · u1) * u1 = ((1,1,0) · (√2/2, 0, √2/2)) / ((√2/2, 0, √2/2) · (√2/2, 0, √2/2)) * (√2/2, 0, √2/2)

= (√2/2) / (1/2 + 1/2) * (√2/2, 0, √2/2) = (√2/2) * (√2/2, 0, √2/2) = (1/2, 0, 1/2)

Step 3: Orthogonalize v2 by subtracting the projection:

u2 = v2 - proj = (1,1,0) - (1/2, 0, 1/2) = (1/2, 1, -1/2)

Step 4: Normalize u2:

u2 = u2 / ||u2|| = (1/2, 1, -1/2) / √(1/4 + 1 + 1/4) = (1/2, 1, -1/2) / √2 = (1/√8, √2/√8, -1/√8) = (1/√8, √2/4, -1/√8)

Step 5: Compute the projection of the third vector onto the subspace spanned by u1 and u2:

v3 = (1,1,2)

proj1 = (v3 · u1) / (u1 · u1) * u1 = ((1,1,2) · (√2/2, 0, √2/2)) / ((√2/2, 0, √2/2) · (√2/2, 0, √2/2)) * (√2/2, 0, √2/2)

= (√2) / (1/2 + 1/2) * (√2/2, 0, √2/2) = (√2) * (√2/2, 0, √2/2) = (1, 0, 1)

proj2 = (v3 · u2) / (u2 · u2) * u2 = ((1,1,2) · (1/√8, √2/4, -1/√8)) / ((1/√8, √2/4, -1/√8) · (1/√8, √2/4, -1/√8))

= (√2) / (1/8 + 2/8 + 1/8) * (1/√8, √2/4, -1/√8) = (√2) * (1/√8, √2/4, -1/√8) = (1, √2/2, -1)

proj = proj1 + proj2 = (1, 0, 1) + (1, √2/2, -1) = (2, √2/2, 0)

Step 6: Orthogonalize v3 by subtracting the projection:

u3 = v3 - proj = (1,1,2) - (2, √2/2, 0) = (-1, 1 - √2/2, 2)

Step 7: Normalize u3:

u3 = u3 / ||u3|| = (-1, 1 - √2/2, 2) / √((-1)^2 + (1 - √2/2)^2 + 2^2) = (-1, 1 - √2/2, 2) / √(3 - 2√2 + 2 + 4) = (-1, 1 - √2/2, 2) / √(9 - 2√2) = (-1/√(9 - 2√2), (1 - √2/2)/√(9 - 2√2), 2/√(9 - 2√2))

Therefore, an orthonormal basis for R3 using the Gram-Schmidt orthogonalization procedure for set B1 is:

u1 = (√2/2, 0, √2/2)

u2 = (1/√8, √2/4, -1/√8)

u3 = (-1/√(9 - 2√2), (1 - √2/2)/√(9 - 2√2), 2/√(9 - 2√2))

(b) Applying the Gram-Schmidt orthogonalization procedure to set B2 = {(2,1,1),(1,0,1),(0,0,2)}:

Step 1: Normalize the first vector:

v1 = (2,1,1)

u1 = v1 / ||v1|| = (2,1,1) / √(2^2 + 1^2 + 1^2) = (2,1,1) / √6 = (2/√6, 1/√6, 1/√6)

Step 2: Compute the projection of the second vector onto the subspace spanned by u1:

v2 = (1,0,1)

proj = (v2 · u1) / (u1 · u1) * u1 = ((1,0,1) · (2/√6, 1/√6, 1/√6)) / ((2/√6, 1/√6, 1/√6) · (2/√6, 1/√6, 1/√6)) * (2/√6, 1/√6, 1/√6)

= (√6/3) / (2/3 + 1/6 + 1/6) * (2/√6, 1/√6, 1/√6) = (√6/3) * (2/√6, 1/√6, 1/√6) = (2/3, 1/3, 1/3)

Step 3: Orthogonalize v2 by subtracting the projection:

u2 = v2 - proj = (1,0,1) - (2/3, 1/3, 1/3) = (1/3, -1/3, 2/3)

Step 4: Normalize u2:

u2 = u2 / ||u2|| = (1/3, -1/3, 2/3) / √((1/3)^2 + (-1/3)^2 + (2/3)^2) = (1/3, -1/3, 2/3) / √(1/9 + 1/9 + 4/9) = (1/3, -1/3, 2/3) / √(6/9) = (1/√6, -1/√6, 2/√6) = (1/√6, -1/√6, √2/√6)

Step 5: Compute the projection of the third vector onto the subspace spanned by u1 and u2:

v3 = (0,0,2)

proj1 = (v3 · u1) / (u1 · u1) * u1 = ((0,0,2) · (2/√6, 1/√6, 1/√6)) / ((2/√6, 1/√6, 1/√6) · (2/√6, 1/√6, 1/√6)) * (2/√6, 1/√6, 1/√6)

= (2√6/3) / (2/3 + 1/6 + 1/6) * (2/√6, 1/√6, 1/√6) = (2√6/3) * (2/√6, 1/√6, 1/√6) = (4/3, 2/3, 2/3)

proj2 = (v3 · u2) / (u2 · u2) * u2 = ((0,0,2) · (1/√6, -1/√6, √2/√6)) / ((1/√6, -1/√6, √2/√6) · (1/√6, -1/√6, √2/√6))

= (2√2/3) / (1/6 + 1/6 + 2/6) * (1/√6, -1/√6, √2/√6) = (2√2/3) * (1/√6, -1/√6, √2/√6) = (√2/3, -√2/3, 2/3√2)

proj = proj1 + proj2 = (4/3, 2/3, 2/3) + (√2/3, -√2/3, 2/3√2) = (4/3 + √2/3, 2/3 - √2/3, 2/3 + 2/3√2) = ((4 + √2)/3, (2 - √2)/3, (2 + 2√2)/3)

Step 6: Orthogonalize v3 by subtracting the projection:

u3 = v3 - proj = (0,0,2) - ((4 + √2)/3, (2 - √2)/3, (2 + 2√2)/3) = (-4/3 - √2/3, -2/3 + √2/3, 2/3 - 2/3√2)

Step 7: Normalize u3:

u3 = u3 / ||u3|| = (-4/3 - √2/3, -2/3 + √2/3, 2/3 - 2/3√2) / √((-4/3 - √2/3)^2 + (-2/3 + √2/3)^2 + (2/3 - 2/3√2)^2)

= (-4/3 - √2/3, -2/3 + √2/3, 2/3 - 2/3√2) / √(16/9 + 8/9 - 8√2/9 + 8/9 + 4/9 + 8√2/9 + 4/9 - 8/9 + 8/9)

= (-4/3 - √2/3, -2/3 + √2/3, 2/3 - 2/3√2) / √(36/9 + 16/9 + 16/9)

= (-4/3 - √2/3, -2/3 + √2/3, 2/3 - 2/3√2) / √(68/9)

= (-√2/√68, √2/√68, 2√2/√68)

= (-1/√17, 1/√17, 2/√17)

Therefore, an orthonormal basis for R3 using the Gram-Schmidt orthogonalization procedure for set B2 is:

u1 = (2/√6, 1/√6, 1/√6)

u2 = (1/√6, -1/√6, √2/√6)

u3 = (-1/√17, 1/√17, 2/√17)

Learn more about orthonormal basis

https://brainly.com/question/32670388

#SPJ11

Exercise 6 If X Is A Continuous Random Variable With A Probability Density Function F(x) = C.sina: 0 (2024)

FAQs

What is the probability density of a continuous random variable X? ›

The probability density function (pdf) f(x) of a continuous random variable X is defined as the derivative of the cdf F(x): f(x)=ddxF(x).

What does the probability density function f x for any continuous random variable x represent? ›

The probability density function, f(x), for any continuous random variable X, represents: the height of the density function at x.

How do we calculate the mean of a continuous random variable with the probability density function? ›

The mean of a continuous random variable is E[X] = μ=∫∞−∞xf(x)dx μ = ∫ − ∞ ∞ x f ( x ) d x and variance is Var(X) = σ2=∫∞−∞(x−μ)2f(x)dx σ 2 = ∫ − ∞ ∞ ( x − μ ) 2 f ( x ) d x .

What is the pdf formula for a continuous random variable? ›

A certain continuous random variable has a probability density function (PDF) given by: f ( x ) = C x ( 1 − x ) 2 , f(x) = C x (1-x)^2, f(x)=Cx(1−x)2, where x x x can be any number in the real interval [ 0 , 1 ] [0,1] [0,1].

What is an example of a continuous random variable? ›

For example, the height of students in a class, the amount of ice tea in a glass, the change in temperature throughout a day, and the number of hours a person works in a week all contain a range of values in an interval, thus continuous random variables.

How do you show that X is a continuous random variable? ›

A random variable X is continuous if possible values comprise either a single interval on the number line or a union of disjoint intervals. Example: If in the study of the ecology of a lake, X, the r.v. may be depth measurements at randomly chosen locations.

What is the probability density function of a random variable X? ›

The probability density function of a random variable X is Px(x)=e−x for x≥0 and 0 otherwise.

What is a continuous probability distribution for a random variable X? ›

The probability distribution of a continuous random variable X is an assignment of probabilities to intervals of decimal numbers using a function f(x), called a density function, in the following way: the probability that X assumes a value in the interval [a,b] is equal to the area of the region that is bounded above ...

What is the density function of a continuous probability distribution? ›

The Probability Density Function

The pdf of a continuous random variable X , is a function f(x) defined such that: Its curve lies on or above the x -axis, i.e. f(x)≥0 f ( x ) ≥ 0 for all x in its range. The area under the entire curve is 1 .

What is a random variable X has a probability mass function? ›

The probability mass function P(X = x) = f(x) of a discrete random variable is a function that satisfies the following properties: P(X = x) = f(x) > 0; if x ∈ Range of x that supports. ∑ x ϵ R a n g e o f x f ( x ) = 1. P ( X ϵ A ) = ∑ x ϵ A f ( x )

What is the density function specifies the probability distribution of a continuous random variable? ›

In probability theory, a probability density function (PDF), density function, or density of an absolutely continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative ...

How to calculate probability of a continuous random variable? ›

If we know the probability density, we can compute the probability for the random variable to be between a a and b: :
  1. P(a≤ξ≤b)=∫ba\w(x)dx P ( a ≤ ξ ≤ b ) = ∫ a b \w ( x ) d x. ...
  2. a=m1(ξ)=∫+∞−∞w(x)xdx a = m 1 ( ξ ) = ∫ − ∞ + ∞ w ( x ) x d x. ...
  3. σ2(ξ)=∫+∞−∞w(x)(x−m1)2dx σ 2 ( ξ ) = ∫ − ∞ + ∞ w ( x ) ( x − m 1 ) 2 d x.
Jan 28, 2021

What is a probability density function with an example? ›

Probability Density Function Example

Say we have a continuous random variable whose probability density function is given by f(x) = x + 2, when 0 < x ≤ 2. We want to find P(0.5 < X < 1). Then we integrate x + 2 within the limits 0.5 and 1. This gives us 1.375.

What is the formula for the mean of a continuous probability distribution? ›

The mean of a continuous uniform distribution between bounds a and b has a simple formula: μ = a + b 2 . This formula makes a large amount of intuitive sense.

What is the probability density of a continuous variable? ›

In probability theory, a probability density function (PDF), density function, or density of an absolutely continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative ...

What is the probability density function of a continuous uniform random variable? ›

A uniform distribution is a continuous random variable in which all values between a minimum value and a maximum value have the same probability. The probability density function is the constant function f(x)=1/(b‐a), which creates a rectangular shape.

How do you find the probability density function? ›

What is the Probability Density Function Formula? We can differentiate the cumulative distribution function (cdf) to get the probability density function (pdf). This can be given by the formula f(x) = dF(x)dx d F ( x ) d x = F'(x). Here, f(x) is the pdf and F'(x) is the cdf.

What is the PMF of a continuous random variable? ›

A continuous random variable takes on an uncountably infinite number of possible values. For a discrete random variable that takes on a finite or countably infinite number of possible values, we determined P ( X = x ) for all of the possible values of , and called it the probability mass function ("p.m.f.").

Top Articles
Latest Posts
Article information

Author: Jerrold Considine

Last Updated:

Views: 6226

Rating: 4.8 / 5 (58 voted)

Reviews: 89% of readers found this page helpful

Author information

Name: Jerrold Considine

Birthday: 1993-11-03

Address: Suite 447 3463 Marybelle Circles, New Marlin, AL 20765

Phone: +5816749283868

Job: Sales Executive

Hobby: Air sports, Sand art, Electronics, LARPing, Baseball, Book restoration, Puzzles

Introduction: My name is Jerrold Considine, I am a combative, cheerful, encouraging, happy, enthusiastic, funny, kind person who loves writing and wants to share my knowledge and understanding with you.