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

\begin{equation} \nu = \left( \iota^1_\ast \pi^1_\ast \mu \right) \ast \left( \iota^2_\ast \pi^2_\ast \mu \right) \, . \end{equation}

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.


1 Answer 1


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 $\mathbb{R}^2$.


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