Discrete Random Variable

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  • #1
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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.
 

Answers and Replies

  • #2
768
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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
768
4
Incidently, are you using Tsokos' new book? The masterpiece of typos?
 
  • #4
24
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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
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Lol yes i am.
 
  • #6
768
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.

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
Ray Vickson
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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
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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
768
4
Yup, must be the same guy. :)
 
  • #10
768
4
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.
 

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