Find the distribution

craig · #1 $Old$ November 2nd, 2009, 01:20 PM

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

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

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

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

Hopefully someone can point me in the right direction with this

Craig

mr fantastic · #2 $Old$ November 2nd, 2009, 04:18 PM

Quote:

Originally Posted by craig $View Post$

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

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

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

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

Hopefully someone can point me in the right direction with this

Craig

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

craig · #3 $Old$ November 2nd, 2009, 04:35 PM

Quote:

Originally Posted by mr fantastic $View Post$

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

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

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

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

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

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

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

Is this what you meant to do?

Thanks for the reply

mr fantastic · #4 $Old$ November 2nd, 2009, 04:39 PM

Quote:

Originally Posted by craig $View Post$

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

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

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

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

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

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

Is this what you meant to do?

Thanks for the reply

Yes.

craig · #5 $Old$ November 2nd, 2009, 04:41 PM

Quote:

Originally Posted by mr fantastic $View Post$

Yes.

Thanks again