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Let $X$ and $Y$ be two independent discrete random variables such that:

P$(y)=1/3,\; y=-1,0,1$

$P(x)=1/2,\; x=2,4$

Let $U=X+Y$

Find the probability mass function of $U.$

I currently have the sum of all $y, x$ of the probability of $X$ multiplied by the probability of $Y$ but it feels wrong :S.

Thanks, any help would be appreciated.

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    Please tell us $P(U = u)$ for a couple of possible values of $u,$ so we can tell if you're on the right track. For example, do $P(U=1)$ and $P(U=3)$ have positive probability? If, so what? – 2017-01-25
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    @BruceET: No I wasn't on the right track. I went back over it with a friend and we found out I was wrong. – 2017-01-25
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    Glad it's straightened out. Thanks for letting us know. – 2017-01-25
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    @BruceET, should I delete this thread or is there a way to mark it as resolved since I got the solution. I have not used this website much so far. – 2017-01-25
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    There is. Not quite sure how. Never did it. Maybe click 'flag' and say what you want to do. // But how about posting your own Answer? Maybe not the details of the solution, but just the results. – 2017-01-25
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    @BruceET, good idea thanks for the help. I answered it myself and I'll accept it once I can confirm it is right. – 2017-01-25

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I believe I was able to answer this myself. I calculated the MGF of X and Y then the MGF of U using those and I used the coefficients from the MGF of U and ended up with:

$P$($U$=1)=1/6

$P$($U$=2)=1/6

$P$($U$=3)=1/3

$P$($U$=4)=1/6

$P$($U$=5)=1/6

If anyone could confirm this for me or tell me why it is wrong or maybe a better way to go about it then that would be greatly appreciated. Thanks!

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    Your answer looks OK. – 2017-01-25