Suppose X is a continuous random variable (has a pdf on the reals), and Y is a mixed random variable (its CDF has atoms). X and Y are independent. Is the sum, Z=X+Y a continuous random variable (i.e., has a pdf on the reals)?
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4$\begingroup$ In thinking about this question I found it useful to contemplate the situation where $Y$ is the most extreme atom possible: a constant. $\endgroup$– whuber ♦Commented Jul 2, 2019 at 18:08
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1$\begingroup$ The "mixed" case literally is that: it's a mixture of a continuous and a discrete distribution. Thus, it suffices to develop intuition for the sum of a mixture distribution and another distribution. The math says the result is a mixture of the sums. $\endgroup$– whuber ♦Commented Jul 2, 2019 at 18:20
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2$\begingroup$ I think you need to be explicit about the independence of $X$ & $Y$ in your question. For example, if $X=C-Y$ for some constant $C$, then it does not matter whether $Y$ is continuous, constant, mixed or discrete. $\endgroup$– AlexisCommented Jul 2, 2019 at 18:24
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2$\begingroup$ The general result follows immediately from the Lebesgue Decomposition Theorem and the definitions. $\endgroup$– whuber ♦Commented Jul 2, 2019 at 19:54
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1$\begingroup$ Both, because--as you demonstrated in your first comment--the decomposition is the heart of the matter and everything else is straightforward to show and understand. $\endgroup$– whuber ♦Commented Jul 5, 2019 at 14:06
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1 Answer
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There is a relatively elementary demonstration that the sum is continuous.
Let $X$ have a probability distribution function $F_X$ with density function $f_X$ and let the distribution function of $Y$ be $F_Y.$ We do not assume $Y$ has a density function. I claim that $X+Y$ has a density function (implying it is absolutely continuous) and its density can be expressed as an expectation,
$$f_{X+Y}(z) = E[f_X(z-Y)] = \int_\mathbb{R} f_X(z-y) \mathrm{d}F_Y(y).$$
To prove this claim, it suffices to show that integrating $f_{X+Y}$ indeed gives the desired probability function for $X+Y.$ The integration is performed by invoking Fubini's Theorem to change the order of the integrals, then changing the variable of integration from $w-y$ to $x,$ and finally expressing a probability in terms of an indicator function $\mathcal{I}.$ The remaining equations are just definitions of distribution functions and expectations as integrals:
$$\eqalign{ \int_{-\infty}^z f_{X+Y}(w)\mathrm{d}w &= \int_{-\infty}^z \int_\mathbb{R} f_X(w-y) \mathrm{d}F_Y(y)\ \mathrm{d}w \\ &= \int_\mathbb{R} \int_{-\infty}^z f_X(w-y) \mathrm{d}w\ \mathrm{d}F_Y(y) \\ &= \int_\mathbb{R} \int_{-\infty}^{z-y} f_X(x) \mathrm{d}x\ \mathrm{d}F_Y(y) \\ &= \int_\mathbb{R} F_X(z-y) \mathrm{d}F_Y(y) \\ &= E[F_X(z-Y)] \\ &= E_Y[\Pr(X \le z-Y)] \\ &= E_Y[E_X[\mathcal{I}(X+Y\le z)]] \\ &= \Pr(X+Y\le z) \\ &= F_{X+Y}(z). }$$
For some intuition, think of adding $X$ to $Y$ as "smearing" every possible value of $Y$ according to the distribution of $X$ or, equivalently, as using $Y$ to weight a mixture of shifted versions of $X.$ In either case it's clear the result will have no atoms because $X$ has no atoms and so (of course) none of its shifted versions have atoms, either, whence no mixture of them will have any atoms.
In this figure, the left panel depicts the density of $X.$ The next panel shows the mass of $Y$ -- this variable has no density. Nevertheless, as shown in the third panel, adding $Y$ to $X$ produces as many continuous components of $X$ as there are spikes in $Y,$ each one scaled by the height of its spike. The density of $X+Y$ is the accumulated height of all these components. Because it is formed from density functions, it too is a density, showing that $X+Y$ follows a continuous distribution even though $Y$ does not.
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$\begingroup$ Quick question: integrating $f_{X+Y}$ shows that $X+Y$ is absolutely continuous? $\endgroup$ Commented Apr 28, 2021 at 18:42
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$\begingroup$ @Taylor Distributions with density functions are absolutely continuous. The Fundamental Theorem of Calculus says that when the integral of $f$ is $F,$ then $f$ is the derivative of $F,$ which makes $f$ a density function for $F.$ $\endgroup$– whuber ♦Commented Apr 28, 2021 at 19:12
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$\begingroup$ Gotcha, so if $X$ and $Y$ are independent absolutely continuous random variables, then $X+Y$ is absolutely continuous with density $f_{X+Y}$? $\endgroup$ Commented Apr 28, 2021 at 19:16
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1$\begingroup$ @Taylor That is a special case of what I have shown here, because it's unnecessary to assume both $X$ and $Y$ have density functions: only one of those variables has to be a.c. for $X+Y$ to be a.c. $\endgroup$– whuber ♦Commented Apr 28, 2021 at 19:19