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  1. #1
    Super Member craig's Avatar
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    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
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  2. #2
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    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 $\displaystyle X_1$ 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 $\displaystyle Y$?
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    Super Member craig's Avatar
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    Quote Originally Posted by mr fantastic View Post
    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
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  4. #4
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    Quote Originally Posted by craig View Post
    $\displaystyle Y = X_1 + X_2$, 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.
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Yes.
    Thanks again
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