# Density of a sum of independent discrete and continuous random vectors

There is a well known fact that a sum of a discrete random variable and a continuous random variable (which are independent) is continuous as well. Is the same statement true for the random vectors?

Specifically, I wanted to calculate a density of a vector $$(X_1, \ldots, X_n) + (U_1, \ldots, U_n)$$ where $$(X_1, \ldots, X_n)$$ is a random variable with values in $$\mathbb{Z}_+^{n}$$ and $$U_1, \ldots, U_n$$ are i.i.d with a uniform distribution over the interval $$(0, 1)$$ and are independent of the vector $$(X_1, \ldots, X_n)$$. I managed to compute the cumulative distribution function of the sum and differentiate it by all the variables which would give me the candidate for a density: $$\mathbb{R}_+^{n} \ni (x_1, \ldots, x_n) \mapsto \mathbb{P}(X_i = \lfloor x_i \rfloor, 1 \leq i \leq n)$$ but I'm not sure how to prove that the integral of that function over $$\mathbb{R}^n_+$$ is equal to 1.

• This "well known fact" is only well known by you… because it's wrong in general.
– Gono
Oct 25, 2019 at 11:57
• @Gono do you have any counterexample in your mind? I'm pretty sure I can prove that for a single-dimensional version using the convolution operation. Oct 25, 2019 at 12:04
• @Gono Under indepndence the fact is true and well known. Oct 25, 2019 at 12:12

If $$X$$ and $$Y$$ are independent random vectors and $$Y$$ has a density then so does $$X+Y$$. This is an easy consequence of Fubini's Theorem.
In measure theoretic language if $$\nu << m$$ then $$\mu *\nu << m$$ for any probability measure $$\mu$$ (wheer $$m$$ is Lebesgue measure.