# Find the distribution

• Nov 2nd 2009, 11:20 AM
craig
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
• Nov 2nd 2009, 02:18 PM
mr fantastic
Quote:

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 $\displaystyle Y$?
• Nov 2nd 2009, 02:35 PM
craig
Quote:

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?

• Nov 2nd 2009, 02:39 PM
mr fantastic
Quote:

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?