anonymous
  • anonymous
Suppose I throw 2 fair dice. Let X be a random variable of the total points from the 2 fair dice thrown. So, the random value X can hold 11 values from 2 to 12: \[X=\{ 2,3,...,12\] As usual in textbooks, there is often another omega in the sample space like this: \[\omega \in \Omega \] In my case, what are the values in my space space of capital Omega? And what are my sampling values?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
In my case, what are the values in my sample space of capital Omega? And what are my sampling values?
ash2326
  • ash2326
\(\omega \)={2,3,4,5,6,7,8,9,10,11,12} I think Ω is {1,2,3,4,5,6,7,8,9,10,11,12}
anonymous
  • anonymous
Are ω the sampling values? Also, since X is a random variable, which is a function by itself, for this: \[X(\omega )=x\] How would I match the values? Suppose I have X(12), what should the output of X(12) be?

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Zarkon
  • Zarkon
X(12) could be 12
anonymous
  • anonymous
When I have X(a)=b, 'a' is the random variable value, which is the total points I get out of the 2 dice, is this right? And what about 'b'? What does it mean for 'b'?
Zarkon
  • Zarkon
a is an element from the sample space...b is a possible value of the random variable
Zarkon
  • Zarkon
You should probably treat it like this... roll two die then the sample space consists of 6*6=36 different ordered pairs let X be the random variable that add the ordered pairs then \(\omega\) is an ordered pair and X is a random variable taking values in {2,3,...,12}
Zarkon
  • Zarkon
so if \(\omega=(2,5)\), \(X(\omega)=7\)
Zarkon
  • Zarkon
X is a measurable function from the sample space to the real line
anonymous
  • anonymous
ohhh..... Thank you so much!! This very clear. Thanks a lot!
anonymous
  • anonymous
Which also means, I cannot write as X(12) because there isn't such a thing. It should be X({6,6})
Zarkon
  • Zarkon
To be completly correct...yes....some times people will say that the experament is to roll two die and add them ... then one could look at X(12) it is better (and correct) to treat the two die rolls as the experament and then get ordered pairs...then let the random variable add them
anonymous
  • anonymous
Now I understand. Thank you so much! It is very clear! :)
Zarkon
  • Zarkon
good ;)

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