If $Z=X+Y$, and the PDFs of $X$ and $Y$ are both functions of a deterministic variable $d$, how can I evaluate the PDF of $z$ while the convolution cannot be used here (due to lack of independence)?
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As discussed in Dilip's response here, you can take the approach of doing direct integration with the bivariate density; however, I want to mention that while this is perhaps the most straightforward and general answer, there are other possibilities.
One other approach that sometimes can be useful (though usually it's effectively the same as the above approach as Dilip Sarwate points out below) is to use bivariate change of variable, for example from $(X,Y)$ to $(U,V) = (X+Y,X-Y)$, say (there are other possibilities for $V$) followed by integrating $V$ out to get the required marginal.
Sometimes the situation may allow the possibility of substantially more convenient approaches, while these approaches I've mentioned are fairly general. If you explain the nature of the dependence between the variables more explicitly, it may be that some simpler approach may help with your specific problem.
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2The transformation $(X,Y) \to (X+Y,X-Y)$ is particularly simple in that the joint pdf of $X+Y$ and $X-Y$ is just the joint pdf of $X$ and $Y$ with the axes rotated by $\pi/4$ and dilated by a factor of $\sqrt{2}$. However, "integrating out" the un-needed $X-Y$ to get the marginal distribution of $X+Y$ leads to essentially the same calculation as the direct integration of the bivariate density that one was hoping to avoid. But there _are_ instances in which this trick does simplify life. For example, if the joint pdf has _circular symmetry_, then $f_{X+Y}(z)=f_X(z/\sqrt{2})$. continued.... – Dilip Sarwate Nov 05 '14 at 14:28
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2...(continuation) An example of this type of calculation (which can be used even for the more general case of $\alpha X+\beta Y$) can be found in [this answer of mine](http://math.stackexchange.com/a/65871/15941) on math.SE. – Dilip Sarwate Nov 05 '14 at 14:31
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@dilip-sarwate: The issue that I have noticed in your solution is that it requires the joint distribution, which I don't have access to it. I only have the marginals, which are convolved if x and y are independent (not the case here). Is my understanding correct? – Pioneer83 Nov 06 '14 at 03:27
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3@Zohair **Any** calculation of the distribution of $X+Y$ requires knowledge of the joint distribution of $X$ and $Y$. You cannot do without it, period. For the case of _independent_ random variables, the joint distribution is the product of the marginal distributions and so the marginal distributions **and the knowledge (or assumption) that $X$ and $Y$ are independent** suffices and gives us the convolution formula. But for dependent random variables, convolving the marginals does not work; you need the joint distribution. – Dilip Sarwate Nov 06 '14 at 03:45