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I know, I can't use convolution. I have two random variables A and B and they're dependent. I need Distributive function of A+B

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    $\begingroup$ If A and B are dependent, then joint distribution of A and B is required to arrive distribution of A+B. $\endgroup$ – vinux Jan 23 '12 at 12:15
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    $\begingroup$ I do not understand your question. What do you know and why can't you use convolution? $\endgroup$ – Xi'an Jan 23 '12 at 13:07
  • $\begingroup$ I know Distributive function of A and B. f A and B are two independent, continuous random variables, then I can find the distribution of Z=A+B by taking the convolution of f(A) and g(B): h(z)=(f∗g)(z)=∫∞−∞f(A)g(z−B)dA But what can I do, when they're not independent ? I am sorry, if this is dumb question. $\endgroup$ – Mesko Jan 23 '12 at 13:39
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    $\begingroup$ It's not a dumb question Mesko, but what people are pointing out is that it needs more information. The answer depends on how $A$ and $B$ fail to be independent. A full description of that is given by the joint distribution of $A$ and $B$, which is what vinux asks. Xi'an is probing a little more delicately but really seeks the same kind of information in order to help you make progress. $\endgroup$ – whuber Jan 23 '12 at 14:08
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As vinux points out, one needs the joint distribution of $A$ and $B$, and it is not obvious from OP Mesko's response "I know Distributive function of A and B" that he is saying he knows the joint distribution of A and B: he may well be saying that he knows the marginal distributions of A and B. However, assuming that Mesko does know the joint distribution, the answer is given below.

From the convolution integral in OP Mesko's comment (which is wrong, by the way), it could be inferred that Mesko is interested in jointly continuous random variables $A$ and $B$ with joint probability density function $f_{A,B}(a,b)$. In this case, $$f_{A+B}(z) = \int_{-\infty}^{\infty} f_{A,B}(a,z-a) \mathrm da = \int_{-\infty}^{\infty} f_{A,B}(z-b,b) \mathrm db.$$ When $A$ and $B$ are independent, the joint density function factors into the product of the marginal density functions: $f_{A,B}(a,z-a)=f_{A}(a)f_{B}(z-a)$ and we get the more familiar convolution formula for independent random variables. A similar result applies for discrete random variables as well.

Things are more complicated if $A$ and $B$ are not jointly continuous, or if one random variable is continuous and the other is discrete. However, in all cases, one can always find the cumulative probability distribution function $F_{A+B}(z)$ of $A+B$ as the total probability mass in the region of the plane specified as $\{(a,b) \colon a+b \leq z\}$ and compute the probability density function, or the probability mass function, or whatever, from the distribution function. Indeed the above formula is obtained by writing $F_{A+B}(z)$ as a double integral of the joint density function over the specified region and then "differentiating under the integral sign.''

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  • $\begingroup$ This is related to my comment and answer on another question dealing with joint distributions a few days ago. $\endgroup$ – Xi'an Jan 23 '12 at 19:27
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Beforehand , I don't know if what i'm saying is correct but I got stuck on the same problem and I tried to solve it in this way:

express the joint distribution using the heaviside step function: $$ f_{A,B}(a,b)=(a+b) H(a,b) H(-a+1,-b+1) $$ or equivalently $$ f_{A,B}(a,b)=(a+b)(H(a)-H(a-1))(H(b)-H(b-1)) $$ Now you can perform the integral without caring about limits of integration.

This is the wolfram rapresentation of the joint : A

Computing the integral I have : B

Plotted : C

That's the function : $$ f(z)=\left\{\begin{matrix} z^2 \qquad \qquad \; \quad for \quad 0\leq z \leq1\\ 1-(z-1)^2 \quad for \quad 1\leq z \leq 2 \\ 0 \qquad \quad \; otherwise \end{matrix}\right. $$ and it's normalized as you can easily check.

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  • $\begingroup$ The question did nt seem to be specific enough about the joint distribution to get an answer. How did you come up with one.? $\endgroup$ – Michael Chernick Feb 13 '18 at 2:02
  • $\begingroup$ +1 for correctly solving the alleged counterexample in @cdlg's answer and showing that the calculations if carried out correctly do give the correct answer, and not the erroneous s results in cdlg's answer. I can't believe that that answer has received two upvotes. $\endgroup$ – Dilip Sarwate Feb 13 '18 at 4:40

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