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Math Help - Find the distribution

  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 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
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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 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?
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  3. #3
    Super Member craig's Avatar
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    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
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  4. #4
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    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.
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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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