What are the Probability Densities for Discrete Random Variable Z?

In summary: But yes, we are looking for the probability mass function for Z, which is given in the attempted solution. In summary, the problem involves finding the probability mass function for a random variable Z, which is equal to X^2 + 1, given the probabilities of the values of X. The solution involves determining the probabilities of Z for each value of X and adding them together for values that are equal to each other, resulting in a probability mass function for Z.
  • #1
iHeartof12
24
0
Let X be a discrete random variable that can assume the values -1, 0,1,2,3,4 with the probabilities 1/6, 1/12, 1/6, 1/4, 1/12, 1/4. Find the probability densities of the following random variables:

a) Z= X^2 + 1

h(y)= f(g^-1(y))


Attempted Solution
X= -1 0 1 2 3 4
Z= 2 1 2 5 10 17

How do i find Pr(Z)?
The book gave an answer of
1/12 Z=1,10
1/4 Z= 5,17
1/3 Z=2
0 elsewhere.
 
Physics news on Phys.org
  • #2
When X= -1, Z=2. So the probability of Z=2 is Pr(x=-1)= 1/6
When X = 0, Z= 1. So the probability of Z=1 is Pr(x=0)= 1/12
and so on

Notice that when X= 1, Z=2 again. Thus, the probability of Z=2 is Pr(x= -1) plus Pr(x=1) which equals 1/3.
 
  • #3
Incidently, are you using Tsokos' new book? The masterpiece of typos?
 
  • #4
Ok thank you.
Let me know if I'm thinking about this correctly.

Z=5 only X=2 can produce this so Pr= 1/4
Z=10 only X=3 can produce this so Pr=1/12
Z= 17 only X=4 can produce this so Pr=1/4.
 
  • #5
Lol yes i am.
 
  • #6
iHeartof12 said:
Ok thank you.
Let me know if I'm thinking about this correctly.

Z=5 only X=2 can produce this so Pr= 1/4
Z=10 only X=3 can produce this so Pr=1/12
Z= 17 only X=4 can produce this so Pr=1/4.

Yes, this is correct.
Do you have a test Thursday? I'm wondering if you are in my class, since this book is so new. 6:30 Tuesdays and Thursdays?
 
  • #7
iHeartof12 said:
Let X be a discrete random variable that can assume the values -1, 0,1,2,3,4 with the probabilities 1/6, 1/12, 1/6, 1/4, 1/12, 1/4. Find the probability densities of the following random variables:

a) Z= X^2 + 1

h(y)= f(g^-1(y))


Attempted Solution
X= -1 0 1 2 3 4
Z= 2 1 2 5 10 17

How do i find Pr(Z)?
The book gave an answer of
1/12 Z=1,10
1/4 Z= 5,17
1/3 Z=2
0 elsewhere.

I hope your book and your instructor do not use the terminology "probability density" in this case. Discrete random variables do not have probability densities; they have distribution functions and probability mass functions, but not densities.

RGV
 
  • #8
ArcanaNoir said:
Yes, this is correct.
Do you have a test Thursday? I'm wondering if you are in my class, since this book is so new. 6:30 Tuesdays and Thursdays?

No I have an exam on Friday. I take it MWF 12:55 but it may be for the same professor our exam covers Chpts. 3 and 4!
 
  • #9
Yup, must be the same guy. :)
 
  • #10
Ray Vickson said:
I hope your book and your instructor do not use the terminology "probability density" in this case. Discrete random variables do not have probability densities; they have distribution functions and probability mass functions, but not densities.

RGV

Oh he uses the term "probability density" all right. This book is a joke.
 

What is a discrete random variable?

A discrete random variable is a type of random variable that can only take on a finite or countably infinite number of values. These values are usually whole numbers and are distinct from one another, hence the term "discrete". Examples of discrete random variables include the number of heads in a coin toss or the number of children in a family.

How is a discrete random variable different from a continuous random variable?

A discrete random variable can only take on a finite or countably infinite number of values, while a continuous random variable can take on any value within a given range. For example, the height of a person is a continuous random variable, as it can take on any value within a certain range (e.g. 5 feet to 6 feet). In contrast, the number of siblings a person has is a discrete random variable, as it can only take on whole number values (e.g. 0, 1, 2, 3...).

What is the probability distribution of a discrete random variable?

The probability distribution of a discrete random variable is a list of all the possible values that the variable can take on, along with their corresponding probabilities. This distribution is often displayed in the form of a table, graph, or formula. It shows the likelihood of each value occurring and can be used to calculate the probabilities of different events involving the random variable.

How do you find the mean of a discrete random variable?

The mean of a discrete random variable is also known as its expected value. To find the mean, you multiply each possible value of the variable by its corresponding probability and then add all of these products together. This calculation can be represented by the formula E(X) = x1p1 + x2p2 + ... + xnpn, where xi is the value of the random variable and pi is its corresponding probability. For example, if you had a random variable representing the number of times a coin lands on heads in a series of 10 coin tosses, the mean would be (0 x 0.5) + (1 x 0.25) + (2 x 0.125) + (3 x 0.0625) + (4 x 0.03125) = 1.5.

What is the difference between a discrete random variable and a discrete probability distribution?

A discrete random variable is a type of variable that can take on a finite or countably infinite number of values, while a discrete probability distribution is a list of all the possible values of a random variable along with their corresponding probabilities. In other words, a discrete random variable is the concept or idea, while a discrete probability distribution is a way to represent that concept or idea.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
944
  • Calculus and Beyond Homework Help
Replies
2
Views
459
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
34
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
975
  • Calculus and Beyond Homework Help
Replies
11
Views
830
  • Calculus and Beyond Homework Help
Replies
4
Views
910
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
6
Views
695
Back
Top