f X ( x ) = { 3 x 2 2 x ∈ [ − 1 , 1 ] 0 e l s e f Y ( y ) = { y 9 y ∈ [ 4 , 5 ] 0 e l s e X and Y are independent RVs, find f X + Y ( z ) {\displaystyle f_{X}(x)={\begin{cases}{\frac {3x^{2}}{2}}&x\in [-1,1]\\0&else\end{cases}}\qquad f_{Y}(y)={\begin{cases}{\frac {y}{9}}&y\in [4,5]\\0&else\end{cases}}\qquad X{\mbox{ and }}Y{\mbox{ are independent RVs, find }}f_{X+Y}(z)}
![{\displaystyle f_{X}(x)={\begin{cases}{\frac {3x^{2}}{2}}&x\in [-1,1]\\0&else\end{cases}}\qquad f_{Y}(y)={\begin{cases}{\frac {y}{9}}&y\in [4,5]\\0&else\end{cases}}\qquad X{\mbox{ and }}Y{\mbox{ are independent RVs, find }}f_{X+Y}(z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56ee9ff5caede25974fdb7b643d3c5d492ba30e8)
![{\displaystyle f_{X}(x)={\frac {3x^{2}}{2}}\cdot 1_{[-1,1]}(x)\qquad f_{Y}(y)={\frac {y}{9}}\cdot 1_{[4,5]}(y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/40d65ca189eaa4c9689fb0e9c14d998b00013420)
仅在 ![{\displaystyle x\in [-1,1]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d536e9c36efe1247b492b0772532bc79c085de2)
时定义，而 
仅在 ![{\displaystyle y\in [4,5]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84fcbe49075e9bea6a4ccbf325402d191b47b9c3)
时定义![{\displaystyle \int _{-\infty }^{\infty }f_{X}(z)f_{Y}(y-z)\delta z=\int _{-\infty }^{\infty }{\frac {3z^{2}}{2}}\cdot 1_{[-1,1]}(z){\frac {y-z}{9}}\cdot 1_{[4,5]}(y-z)\delta z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c56fd760e2aa344e183485cda02bce2a0ebc90a8)
![{\displaystyle {\frac {3}{18}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z={\frac {1}{6}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bc18f9ab9a274544da4738f01871b0aa7067033)
的指示函数因式分解到积分边界![{\displaystyle {\frac {1}{6}}\int _{-1}^{1}z^{2}(y-z)\cdot 1_{[4,5]}(y-z)\delta z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/499d56e4e7b2585dd1c107921fc1e04443741d64)
![{\displaystyle {\frac {1}{6}}\int _{-1}^{1}\left(z^{2}(y-z)\right)\cdot 1_{[y-5,y-4]}(z)\delta z={\frac {1}{6}}\int _{-1}^{1}\left(z^{2}y-z^{3}\right)\cdot 1_{[y-5,y-4]}(z)\delta z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/446df2d15b1b05c4a6c2fbb76a9bd2890a2bf624)
时，积分没有边界，因为 
，因此上界将小于 
，这将是 
。
时，积分的边界在 和 
之间，因为 将至少为 ，但小于 ![{\displaystyle \int _{a}^{b}(\cdot )1_{[c,d]}\delta z\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6c997285efd8ef468952e76f18970db9c22c098)
，您将有 
