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Let $f_V$ be a pdf of random variable $V$ where \begin{align} V=U+Z \end{align} and where $U$ and $Z$ are independent and $Z$ is Gaussian. Moreover, suppose that $|U| \le A$.

Can we find the upper bound on the number of modes of $V$?

Note that the the pdf $V$ is given by \begin{align} f_V(v)=E\left[\frac{1}{\sqrt{2\pi}}e^{-\frac{(v-U)^2}{2}}\right] \end{align}

I am actually not interested in a very tight bound. I more interested in the methods that can be used to produce such bounds.

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The pdf $f_V$ is a mixture of the family $\F_A:=(f_a)_{a\in[-A,A]}$ of the normal pdfs $f_a$ with mean $a$ and variance $1$. By Theorem 6 on p. 2129 of Kemperman, for each natural $s$, the pdf $f_V$ will have at most $s$ modes (or, more precisely, modal intervals) iff any mixture of any $2s$ members of the family $\F_A$ has at most $s$ modal intervals.

This theorem is a generalization of Theorem 4 on p. 2128 of Kemperman, which implies that all mixtures of a family of pdfs are unimodal iff any mixture of any two members of the family is unimodal.

Further, by Remark 1 on p. 2133 of Kemperman, our pdf $f_V$ will be unimodal for all admissible $U$ iff for all $a,b,x$ such that $-A\le a<x<b\le A$ we have \begin{equation} f'_a(x)f''_b(x)\ge f'_b(x)f''_a(x), \end{equation} which can be rewritten as $(a-x)(b-x)\ge-1$ for all such $a,b,x$, which is equivalent to $A\le1$.

The case of more modes than $1$ appears much more difficult analytically.

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