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I have 2 random variables X and Y with the distribution table:

$$ X,Y{\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}} \begin{pmatrix} -1 & 0 & 1 \\ 0.25 & 0.5 & 0.25 \\ \end{pmatrix} $$

I have to find $$ P(X+Y<0) $$

Now I don't have an idea on how to find the distribution table of X+Y if that's necessary.

For now I know that $$ E(X+Y)= E(X)+E(Y) = 0 $$ and $$ V(X) = 0.5 $$

So

$$ P(X+Y<0) = P(X=-1,Y=-1)+P(X=-1,Y=0)+P(X=0,Y=-1) $$

Can you please point me in the right direction from what should I find now? My thoughts where that I need to find the distribution table of X+Y in order to find those 3 probabilities, but I don't know how to do it.

Thanks!

  • 1
    Are $X$ and $Y$ independent? If so then apply $P(X=x,Y=y)=P(X=x)P(Y=y)$. – 2017-02-08
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    I thought about that, but I don't know if they are independent or not, it's not specified. – 2017-02-08
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    If they aren't independent, you usually can't say anything about $X+Y < 0$. – 2017-02-08
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    If nothing more is known then the problem cannot be solved. If e.g. $Y=X$ then you find $P(X+Y<0)=P(2X<0)=P(X<0)=0.25$ and if $Y=-X$ (note that $X$ and $-X$ have equal distribution) you find $P(X+Y<0)=P(0<0)=0$. Both cases (and more) are possible. – 2017-02-08

2 Answers 2

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You can not find this probability without knowledge on joint distribution.

For independent $X$ and $Y$ this probability equals $\frac{5}{16}$. For dependent random variables the answer can be quite different. Say, for $X=-Y$, $P(X+Y<0)=P(0<0)=0$. You can obtain the other values of probability for other form of dependence.

0

$P = P(X + Y < 0 | X = -1)P(X = -1) + P(X + Y < 0 | X = 0)P(X = 0) + P(X + Y < 0 | X = 1)P(X = 1)$

$ = (.25 + .5)(.25) + (.25)(.5) + 0 = 5/16$

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    As the comments show, the OP is not willing to assume that $X,Y$ are independent, so computations of this form are not valid. – 2017-02-08
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    @lulu - If it's an academic question, then that's my best guess at what the answer is likely to be - his given values of E and var don't give him any info about independence that I can see – 2017-02-08