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Let $R_1$ and $R_2$ denote two real-valued discrete random variables.

The probability convolution is the probability distribution for the sum $R_3:=R_1+R_2$ when $R_1$ and $R_2 $ are independent. For example, if $p(R_1=3)=0.2;p(R_1=0)=0.8$ and $p(R_2=2)=0.6;p(R_2=1)=0.4$. Then $$p(R_3=3+2)=0.2*0.6,$$ $$p(R_3=3+1)=0.2*0.4,$$ $$p(R_3=0+2)=0.8*0.6,$$ $$p(R_3=0+1)=0.8*0.4.$$

I wonder if any one could help writing the expression of the probability distribution for $R_3$ when $R_1$ and $R_2$ are dependent?

Besides, how can we write the expression neatly such that the common values in the sum are collected into one value with a single probability?

Emma
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  • Even your form for independent variables is not quite right: you have to collect all cases with a common value of $\lambda x_i + (1-\lambda)y_j$ into one value with a single probability. For instance, with $R_1=R_2=(0,1/2; 1;1/2)$ and $\lambda=1/2$ you have $R_3=(0,1/4;1/2,1/2;1,1/4).$ Note there are only three possible values, not four. – whuber Mar 27 '22 at 03:34
  • You are correct. Thanks. This is also part of my question. I am struggling with how to express the probability mass function correctly in the intended format. – Emma Mar 27 '22 at 11:12
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    Perhaps https://stats.stackexchange.com/questions/331973 will help you. When the variables take on lattice values (that is, all values are integral multiples of some common quantity), then the probability generating function is a convenient way to express the distribution in this form. Are these suggestions consistent with what you're looking for? – whuber Mar 27 '22 at 14:00

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