Random Variables: Proving Same Probability Distribution & Finding $X+Y$

[Julio1]

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 $X+Y.$

Spoiler

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!

 

[I like Serena]

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 $X+Y.$

Spoiler

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?

 

[Julio1]

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 !

 

[I like Serena]

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 !

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?

 

[Julio1]

OK, thanks!

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

 

[I like Serena]

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 $X = 1$.

 

[Julio1]

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 $\dfrac{1}{3}$?

 

[I like Serena]

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! ;)

 

[Julio1]

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.

 

[I like Serena]

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.

Ah. But we still have the second part of the question, which sort of addresses that.

 

[Julio1]

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?

 

[I like Serena]

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.