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I have a question that I encountered while reading my notes. Suppose we have two independent variables

\begin{align*} &P_X(dx) = e^{-x} \mathbb{1}_{\mathbb{R}_+}dx \\ &P_Y = \frac{1}{2}\sum\limits^{\infty}_{k=0} 2^{-k} \delta_k \end{align*}

I need to calculate $\mathbb{P}({X+Y \leq 3})$.

What I have tried: According to the examples with continuous variables, I thought that there's two way to calculate it. First of all we need to find $P_{(X,Y)}$ which is easy then we can either take measure on set $G=\{(x,y)\in \mathbb{R}^2 : x+y\leq 3\}$. Or calculate $P_{f(X,Y)}$, where $f(X,Y)=X+Y$. Both time I either fail to calculate the integrals or their bounds (I think since I don't know the answer, the problematic part is mixing discrete and continuous part). It would be nice if someone could show how to do it either way, so I could know what I do wrong.

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Hint: You can partiton the event by the values of $Y$, i. e. $P(X + Y \le 3) = \sum_{k=0}^\infty P(X+ Y \le 3, Y = k)$. And $P(X+ Y \le 3, Y = k) = P(X \le 3 -k, Y = k)$. –  martini May 16 '12 at 15:51
This looks a lot like homework, and if so, please add the homework tag. You might want to delete the measure theory tag too since the use of formal measure theory here is really using a bomb to kill a fly. The probability that you want is most easile computed via the law of total probability as $$P\{X+Y \leq 3\} = \sum_{i=0}^3 P\{X \leq 3-i \mid Y = i\}P\{Y = i\}$$ –  Dilip Sarwate May 16 '12 at 15:52
You both got with the same answer about the total probability. But answer isn't actually the most important thing for me. I am interested to know the formal calculus how to calculate it, using pushforward measure. –  user974514 May 16 '12 at 20:42
In the present case, to compute the full distribution of $X+Y$ and to deduce from this distribution the value of $\mathbb P(X+Y\leqslant3)$ should be strongly discouraged. –  Did May 18 '12 at 6:41

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