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This question belongs to a problem set of MIT OCW probabilistic systems analysis and applied probability.

Random variables X and Y have the joint PMF(probability mass function)

P$_{X,Y}$ (x, y)=$\frac1 {72}$(x$^2$ + y$^2$), if x ∈ {1, 2, 4} and y ∈ {1, 3},

P$_{X,Y}$ (x, y) = 0, otherwise.

Ques. What is var(X+Y)? (var denotes variance)

My doubt: When I do it using this formula E[(X+Y)$^2$]−(E[X+Y])$^2$ , I get answer as $\frac {2188} {72} -\frac {256} 9= \frac {35} {18}$. This matches with solution provided.

But when I solve it using this formula E[(X+Y - $\mu_{X+Y}$)$^2$], I get answer as $\frac {115} {72}$. -->where $\mu_{X+Y}$ denotes expectation of (X+Y)

I have checked multiple times for calculation mistakes. I believe I have made logical error that is why I am asking here.

This is my solution using this formula E[(X+Y - $\mu_{X+Y}$)$^2$] :- (at the end answer comes out to be $\frac {1035} {9*72}$ ) What errors have I made? enter image description here

Ayush
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  • There must be some arithmetic or algebraic error. The two formulas are easily shown to be identical. Let $Z=X+Y$. Var$(Z)=E((Z-E(Z))^2)=E(Z^2-2ZE(Z)+E(Z)^2)$$=E(Z^2)-2E(Z)E(Z)+E(Z)^2=E(Z^2)-E(Z)^2$ – herb steinberg Jul 08 '21 at 21:38
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    I think your final line is $0.308641975 + 0.24691358 + 0.37808642 +0.020061728+ 0.026234568+ 0.964506173 $ $= 1.944444444$ which is $\frac{35}{18}$ – Henry Jul 08 '21 at 22:00
  • I feel sorry to ask this stupid question. It turns out there was nothing but calculation mistake which I was doing in every attempt(~20 times). Next time I will wait more before asking here. Thank you for your valuable time. – Ayush Jul 09 '21 at 05:40

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