Question

(iv) Let X be exponentially distributed with parameter 1 and let Y be uniformly distributed in the interval [0, 1]. Using convolution, find the probability distribution function of

Answer 1

$CDF~of~Z~is\\ F_Z(z)=P(|X-Y|\leq z)=P(-z\leq X-Y\leq z)......(1)\\ The~pdf~of~U=X-Y~is\\ f_U(u)=\int_{-\infty}^\infty f_X(u&plus;y)f_Y(y)dy\\ where,~f_X(u&plus;y)=\exp(-u-y)~where,~u&plus;y>0~or~y>-u\\ =0~otherwise\\ f_Y(y)=1~if~0\leq y\leq 1\\ =0~otherwise\\ Hence~if~-1<u<0~then~f_X(u&plus;y)f_Y(y)=\exp(-u-y)~if~-u\leq y\leq 1\\ if~u>0~then~f_X(u&plus;y)f_Y(y)=\exp(-u-y)~if~0\leq y\leq 1\\ Hence~pdf~of~U~is\\ f_U(u)=\int_{-u}^1\exp(-u-y)dy\\ =\exp(-u)\left[-\exp(-y) \right ]^1_{-u}=\exp(-2u)-\exp(-u-1),~-1<u<0\\ =\int_0^1\exp(-u-y)dy=\exp(-u)\left[-\exp(-y) \right ]^1_{0}\\ =\exp(-u)-\exp(-u-1),~u>0\\$

$Hence~F_Z(z)=\int_{-z}^0\left[\exp(-2u)-\exp(-u-1) \right ]du&plus;\int_0^z\left[\exp(-u)-\exp(-u-1) \right ]du\\ =\frac{1}{2}&plus;\frac{1}{2}\exp(2z)&plus;\exp(z-1)-\exp(-z)-\exp(-z-1)~if~0<z<1\\ =\int_{-1}^0\left[\exp(-2u)-\exp(-u-1) \right ]du&plus;\int_0^z\left[\exp(-u)-\exp(-u-1) \right ]du\\ =\frac{1}{2}&plus;\frac{1}{2}\exp(2)&plus;\exp(-2)-\exp(-z)&plus;\exp(-z-1)~if~z\geq 1\\ PDF~of~Z~is\\ f_Z(z)=\frac{\mathrm{d} F_Z(z)}{\mathrm{d} z}=\left\{\begin{matrix} \exp(2z)&plus;\exp(z-1)&plus;\exp(-z)&plus;\exp(-z-1)~if~0<z<1\\ \exp(-z)-\exp(-z-1)~if~z\geq 1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 0~otherwise~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \end{matrix}\right.$