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Random Variables

Julio

Member
Feb 14, 2014
71
Let $\Omega=\{\omega_1,\omega_2,\omega_3\}$ an sample space, $P(\omega_1)=P(\omega_2)=P(\omega_3)=\dfrac{1}{3},$ and define $X,Y$ and $Z$ random variables, such that

$X(\omega_1)=1, X(\omega_2)=2, X(\omega_3)=3$

$Y(\omega_1)=2, Y(\omega_2)=3, Y(\omega_3)=1$

$Z(\omega_1)=3, Z(\omega_2)=1, Z(\omega_3)=2.$

Show that these three random variables have the same probability distribution. Find the probability distribution of $X+Y.$

Hi !, my name is Julio. I'm from Chile, my english isn't good, sorry, but try to do the best I can. Some indication for the problem?, thanks!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Let $\Omega=\{\omega_1,\omega_2,\omega_3\}$ an sample space, $P(\omega_1)=P(\omega_2)=P(\omega_3)=\dfrac{1}{3},$ and define $X,Y$ and $Z$ random variables, such that

$X(\omega_1)=1, X(\omega_2)=2, X(\omega_3)=3$

$Y(\omega_1)=2, Y(\omega_2)=3, Y(\omega_3)=1$

$Z(\omega_1)=3, Z(\omega_2)=1, Z(\omega_3)=2.$

Show that these three random variables have the same probability distribution. Find the probability distribution of $X+Y.$

Hi !, my name is Julio. I'm from Chile, my english isn't good, sorry, but try to do the best I can. Some indication for the problem?, thanks!
Hi Julio! Welcome to MHB! :)

Hint: What is the probability distribution of X?
That is, what is the probability that X=1, X=2, respectively X=3?
 

Julio

Member
Feb 14, 2014
71
Thanks I Like Sirena.

The probability distribution of an random variable $X$ is $P(X=x)=p(x)$ if $X$ is discrete, and $P(X=x)=f(x)$ if $X$ is continuous.

But really don't understand how solve the problem, help me please ! :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Thanks I Like Sirena.

The probability distribution of an random variable is $P(X=x)=p(x)$ if $X$ is discrete, and $P(X=x)=f(x)$ if $X$ is continuous.

But really don't understand how solve the problem, help me please ! :confused:
So the question is what is $P(X=1)$?
Since $P(\omega_1)=\frac 1 3$ and $X(\omega_1)=1$, it follows that $P(X=1) = \frac 1 3$.
Similarly $P(X=2)=P(X=3)=\frac 1 3$.

How about Y? Does it have the same distribution?
 

Julio

Member
Feb 14, 2014
71
OK, thanks!

The query is how know that $\omega_1=1$?, I don't is notation used.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
OK, thanks!

The query is how know that $\omega_1=1$?, I don't is notation used.
You don't.
What you do know, is that $\omega_1$ is a possible outcome with probability $1/3$.
And furthermore, if the outcome is $\omega_1$, that $X = 1$.
 

Julio

Member
Feb 14, 2014
71
Thanks, then

$P(Y=1)=P(Y=2)=P(Y=3)=P(\omega_n)=\dfrac{1}{3}.$

Truth?

Then, the random variables have the same probability distribution because all are equal $\dfrac{1}{3}$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Thanks, then

$P(Y=1)=P(Y=2)=P(Y=3)=P(\omega_n)=\dfrac{1}{3}.$

Truth?

Then, the random variables have the same probability distribution because all are equal $\dfrac{1}{3}$?
Yep! ;)
 

Julio

Member
Feb 14, 2014
71
One question, what I find strange is that these random variables take these different values​​, and in the end, your probabilities give the same result. :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
One question, what I find strange is that these random variables take these different values​​, and in the end, your probabilities give the same result. :confused:
Ah. But we still have the second part of the question, which sort of addresses that.
 

Julio

Member
Feb 14, 2014
71
Ah. But we still have the second part of the question, which sort of addresses that.
Oh truth !, I had forgotten. Good, I think that is only add. Namely,

$X(\omega_1)+X(\omega_2)+X(\omega_3)=1+2+3=6,$

$Y(\omega_1)+Y(\omega_2)+Y(\omega_3)=2+3+1=6.$

Thus, $(X+Y)(\omega_n)=6+6=12.$

Okay?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Oh truth !, I had forgotten. Good, I think that is only add. Namely,

$X(\omega_1)+X(\omega_2)+X(\omega_3)=1+2+3=6,$

$Y(\omega_1)+Y(\omega_2)+Y(\omega_3)=2+3+1=6.$

Thus, $(X+Y)(\omega_n)=6+6=12.$

Okay?
I'm afraid not.
The first possible outcome is $\omega_1$ with probability $\frac 1 3$.
With $X(\omega_1)=1$ and $Y(\omega_1)=2$, that means that $X+Y=1+2=3$.
So $P(X+Y=3) = \frac 1 3$.

The interesting part is that even though X and Y have the same probability distribution, we get to be surprised by the probability distribution of the sum X+Y.