# Is the convolution of projected probability measures equal to the original measure?

Let $$V$$ be a real vector space, $$\mathcal{A}$$ a $$\sigma$$-algebra over $$V$$ and $$\mu$$ a probability measure on $$\mathcal{A}$$. Suppose there are real vector spaces $$V_1$$, $$V_2$$ such that $$V = V_1 \oplus V_2$$. We then have canonical projections $$\pi^1 : V \to V_1$$ and $$\pi^2 : V \to V_2$$ as well as canonical inclusion maps $$\iota^1 : V_1 \to V, v_1 \mapsto \left( v_1, 0 \right) \in V_1 \oplus V_2 = V$$ and $$\iota^2$$ analogously.

We may now construct the convolution of the iterated pushforward measures as

$$$$\nu = \left( \iota^1_\ast \pi^1_\ast \mu \right) \ast \left( \iota^2_\ast \pi^2_\ast \mu \right) \, .$$$$

Do we have $$\mu = \nu$$? Intuitively, this should be the case but I seem to be unable to prove it. If this is false, please provide a counterexample and/or some criteria when one might expect it to be true.

This is false in general. Let $$V = \mathbb{R}^2$$ and $$\mu$$ be (normalized) $$1$$-dimensional Lebesgue measure on the line segment $$\{(t,t) : 0 \leq t \leq 1 \}$$. Then $$\pi^1_* \mu$$ is Lebesgue measure on the $$x$$-axis restricted to $$[0,1]$$ and $$\pi^2_*\mu$$ is Lebesgue measure on the $$y$$-axis restricted to $$[0,1]$$. But the convolution of these two measures is $$2$$-dimensional Lebesgue measure on the unit square in $$\mathbb{R}^2$$.