Thread: Find the distribution

Hi, hopefully someone can help me with this question, not sure what it is I'm supposed to do exactly.

Let $\displaystyle X_1$ and $\displaystyle X_2$ be independent random variables with common distribution defined by:

$\displaystyle X =$
0, with probability $\displaystyle \frac{1}{8}$
1, with probability $\displaystyle \frac{3}{8}$
2, with probability $\displaystyle \frac{1}{2}$

Find the distribution of the sum of $\displaystyle X_1 + X_2$

Hopefully someone can point me in the right direction with this

Craig

Originally Posted by craig

Hi, hopefully someone can help me with this question, not sure what it is I'm supposed to do exactly.

Let $\displaystyle X_1$ and $\displaystyle X_2$ be independent random variables with common distribution defined by:

$\displaystyle X =$
0, with probability $\displaystyle \frac{1}{8}$
1, with probability $\displaystyle \frac{3}{8}$
2, with probability $\displaystyle \frac{1}{2}$

Find the distribution of the sum of $\displaystyle X_1 + X_2$

Hopefully someone can point me in the right direction with this

Craig

Let $\displaystyle Y = X_1 + X_2$. What are the possible values of $\displaystyle Y$? What values of $\displaystyle X_1$ and $\displaystyle X_2$ give those values? Therefore, what is the probability of those values of $\displaystyle Y$?

Originally Posted by mr fantastic

Let $\displaystyle Y = X_1 + X_2$. What are the possible values of $\displaystyle Y$? What values of $\displaystyle X_1$ and $\displaystyle X_2$ give those values? Therefore, what is the probability of those values of $\displaystyle Y$?

$\displaystyle Y = X_1 + X_2$, therefore $\displaystyle Y$ can take the values 0..4.

$\displaystyle P(Y=0) = \frac{1}{8}\times\frac{1}{8} = \frac{1}{64}$

$\displaystyle P(Y=1) = 2(\frac{3}{8}\times\frac{1}{8}) = \frac{3}{32}$

$\displaystyle P(Y=2) = \frac{17}{64}$

$\displaystyle P(Y=3) = \frac{3}{8}$

$\displaystyle P(Y=4) = \frac{1}{4}$

Is this what you meant to do?

Thanks for the reply

Originally Posted by craig

$\displaystyle Y = X_1 + X_2$, therefore $\displaystyle Y$ can take the values 0..4.

$\displaystyle P(Y=0) = \frac{1}{8}\times\frac{1}{8} = \frac{1}{64}$

$\displaystyle P(Y=1) = 2(\frac{3}{8}\times\frac{1}{8}) = \frac{3}{32}$

$\displaystyle P(Y=2) = \frac{17}{64}$

$\displaystyle P(Y=3) = \frac{3}{8}$

$\displaystyle P(Y=4) = \frac{1}{4}$

Is this what you meant to do?

Thanks for the reply

Yes.

Originally Posted by mr fantastic

Yes.

Thanks again