imagine a random experiment, where first some number $u$ is drawn uniformly on $[c-\varepsilon,c+\varepsilon]$ for $c>0$ and $0<\varepsilon<c$. Next, a $N(u,\sigma^2)$-distributed random variable $Y$ is generated (meaning that we use the $u$ of the first step as mean and assume some known variance $\sigma^2>0$ for the normal distribution). Now, I am interested in the density of $Y$. How can one derive this density? Should I compute the joint density of $u$ and $Y$?

Thanks a lot!


1 Answer 1


Density of $Y$:

\begin{align} f_Y(y) &= \int_{u=c-\epsilon}^{c+\epsilon} f_{Y|U}(y\mid u)f_U(u)\;du \\ &= \int_{u=c-\epsilon}^{c+\epsilon} \dfrac{1}{2\epsilon}\dfrac{1}{\sqrt{2\pi\sigma^2}}\;e^{-\dfrac{1}{2}\left(\dfrac{y-u}{\sigma}\right)^2}\;du. \\ \end{align}

Let $z=(u-y)/\sigma\;$ so that $dz=du/\sigma$. Then,

\begin{align} f_Y(y) &= \dfrac{1}{2\epsilon}\int_{z=(c-\epsilon-y)/\sigma}^{(c+\epsilon-y)/\sigma} \dfrac{1}{\sqrt{2\pi}}\;e^{-\frac{1}{2}z^2}\;dz \\ & \\ &= \dfrac{1}{2\epsilon}\left[ \Phi\left(\dfrac{c+\epsilon-y}{\sigma}\right) - \Phi\left(\dfrac{c-\epsilon-y}{\sigma}\right)\right]. \end{align}


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