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Let $X$ be a real valued random variable with distribution $F$ and $V \sim$ $N(0,1)$ and $X,V$ are independent. For any map $f$ the density of $Y=f(X)+V$ is given by $p_Y(y)=\int\phi(y-f(x))dF(x)$ where $\phi(x)=(2\pi e^{x^2})^{-1/2}$.

I am not able to proceed further, what does it mean to integrate with respect to a distribution function. Any suggestions/hints would be helpful.

Dinesh
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  • You cannot evaluate the integral without knowing what $F$ is. – Kavi Rama Murthy Oct 04 '21 at 09:14
  • Yes I agree. But why does $p_Y$ take this form? – Dinesh Oct 04 '21 at 09:16
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    If $X$ has a density $h$, you could replace $dF(X)$ by $h(x) dx$. (Not using $f$ for the density since you are already using $f$ for another meaning.) Otherwise, check out https://math.stackexchange.com/questions/380785/what-does-it-mean-to-integrate-with-respect-to-the-distribution-function – Jukka Kohonen Oct 04 '21 at 09:17
  • The formual itself is wrong in general. Do you know that $X$ and$V$ are independent? If that is so then the result folows by Fubini's Theorem. – Kavi Rama Murthy Oct 04 '21 at 09:18
  • @KaviRamaMurthy Pardon me. $X$ and $V$ are independent. No special assumption on $f$. They are any Borel maps. – Dinesh Oct 04 '21 at 09:21
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    What level of rigour do you require? I have an answer but it I need to swap a derivative and an integral. I expect this will be justified by DCT, but I don't have time right now to think about the details – JackT Oct 04 '21 at 09:31
  • Since the question is "what does it mean to integrate with respect to a distribution function", would not https://math.stackexchange.com/questions/380785/what-does-it-mean-to-integrate-with-respect-to-the-distribution-function already answer the question? – Jukka Kohonen Oct 04 '21 at 09:34

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If $F_Y(y) = \mathrm{Pr} (Y\leqslant y)$ then \begin{align*} F_Y(y) &= \mathrm{Pr}(f(X)+V \leqslant y). \end{align*} Since $X$ and $V$ are independent, $$\mathrm{Pr}(f(X)+V \leqslant y) = \int_{-\infty}^\infty \int_{-\infty}^{y-f(x)} \phi (v) \, dv\,dF(x). $$ Then* \begin{align*} f_y(y) &= F_Y'(y) \\ &= \int_{-\infty}^\infty \phi (y-f(x)) \, dF(x) \end{align*} as required.

*In order to swap the integral and the derivative here I expect you will need to use dominated convergence. This might require some assumptions on $f$.

Edit: Let $$\psi(x,y) = \int_{-\infty}^{y-f(x)} \phi (v) \, dv = \int_{-\infty}^{y} \phi (v-f(x)) \, dv $$ by a change of variables. Since $v \mapsto \phi (v-f(x))$ is continuous for each $x$, if the lower bound on the above integral was finite then the Fundamental Theorem of Calculus would directly imply that $$\frac{\partial}{\partial y} \psi (x,y) =\phi(y-f(x)). \tag{$\ast$}$$ However, the lower bound is not finite, but $\phi$ decays rapidly so we might expect that $(\ast)$ is still true. Indeed, for $h>0$, \begin{align*} \frac{\psi(x,y+h)-\psi(x,y)}{h} &= \frac1h \int_{y}^{y+h}\phi (v-f(x)) \, dv \\ &= \frac1h \int_0^{h}\phi (v+y-f(x)) \, dv. \end{align*} Since $\phi$ is Lipschitz continuous (its derivative is bounded) it follows that \begin{align*} \bigg \vert \frac{\psi(x,y+h)-\psi(x,y)}{h} - \phi(y-f(x))\bigg \vert &\leqslant \frac1h \int_0^{h} \big \vert \phi (v+y-f(x)) -\phi (y-f(x)) \big \vert \, dv \\ &\leqslant \frac{C}h\int_0^{h} \vert v \vert \, dv \\ &= Ch \to 0 \end{align*} as $h \to 0$. Hence, $(\ast)$ is true and this limit is uniform in $x$. Thus, \begin{align*} F_Y'(y) &=\lim_{h\to0}\int_{-\infty}^\infty \frac1h \int_{y-f(x)}^{y+h-f(x)} \phi (v) \, dv\,dF(x)\\ &= \int_{-\infty}^\infty \lim_{h\to0}\frac1h \int_{y-f(x)}^{y+h-f(x)} \phi (v) \, dv\,dF(x) \\ &= \int_{-\infty}^\infty \psi_y(x,y) \,dF(x) \\ &= \int_{-\infty}^\infty \phi (y-f(x)) \, dF(x) \end{align*}

JackT
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