I am given two independent random variables $X$ and $Y$ with density functions $$ f_X(u) = 2u \cdot \chi_{(0,1)}(u),\qquad f_Y(v) = 2v\cdot \chi_{(0,1)}(v), $$ where $\chi_A(t)= 0$ if $t\not\in A$ and $\chi_A(t) = 1$ if $t\in A$. I would like to determine directly from the joint PDF $f_{X,Y}$ the PDF of the random variable $Z:=X+Y$, and also to determine the joint PDF of the vector $(Z,W)$ where $W:=\min\{X,Y\}$. The first part isn't hard since I introduce a random variable $U:=Y$ in order to obtain the inverse of the measurable transformation, that is: $$ \{z = x+y; u = y \leadsto \{x = z-u; y = u $$ so since the Jacobian of the inverse is $1$ I would get $$ f_Z(z) = \int_{-\infty}^\infty f_{Z,U}(z,u) du = \int_{-\infty}^\infty f_X(z-u) f_Y(u) du. $$ where the integrand is 0 unless $0 \leq z-u< 1 \implies z-1 < u\leq z$ and $0< u < 1$. This gives me $$ f_{Z}(z) = \left\{ \begin{array}{lll} \int_0^z 4(z-u)u du &= \frac{2}{3} z^3 & \text{if } 0 < z < 1,\\ \int_{z-1}^1 4 (z-u)u du &= -\frac{2}{3} (z^3 -6z + 4) & \text{if } 1 \leq z < 2\\ 0 & & \text{otherwise.}\end{array} \right. $$ I've checked that $\int_{-\infty}^\infty f_Z(z) dz = 1$. (Is this correct?, is there a shorter approach?). However, for the second part I'm having doubts because of the definition $\min\{X,Y\}$. A first attempt I've considered is using $$ F_{W}(w) = \mathrm P(W\leq w) = \mathrm P(\min\{X,Y\} \leq w\} = 1- \mathrm P(\min\{X,Y\} > w) = 1-\mathrm P(X > w, Y > w) $$ $$ = 1- \mathrm P(X > w) \mathrm P(Y > w) = 1-(1- F_X(w)) (1-F_Y(w))=\cdots $$ since $\min$ is not differentiable. I would appreciate some help, only hints if you prefer. Thanks in advance.


Your first calculation is correct but a bit longer than necessary. There is no need to introduce a joint density of $Z$ and $U$ etc. Just write $$F_Z(z) = P\{X+Y \leq z\} = \int_{-\infty}^\infty \int_{-\infty}^{z-y}f_X(x)f_Y(y)\,\mathrm dx\, \mathrm dy$$ and differentiate with respect to $z$ to get $$f_Z(z) = \int_{-\infty}^\infty f_X(z-y)f_Y(y)\,\mathrm dy.$$

For the joint density of $Z$ and $W = \min\{X,Y\}$, you can try to find the joint CDF of $Z$ and $W$ and differentiate to get the joint pdf. Begin by finding the region of the $z$-$w$ plane that is the_support_ of $f_{Z,W}(z,w)$.

  • $\begingroup$ Thanks for your answer. However, I am wondering what steps you have followed in order to get those results. For example, "Just write $F_Z(z)=\mathrm P\{X+Y\leq z\}=\int_{-\infty}^\infty \int_{-\infty}^{z-y}...$" is not clear to me. Also, for the CDF in 2nd part, I have $F_{Z,W}(z,w) = P(X+Y\leq z, \min\{X,Y\}\leq w)$, but what else?. The support is $(0,2)\times(0,1)$. $\endgroup$ – V. Galerkin Jan 18 '14 at 15:19
  • $\begingroup$ @V.Galerkin Sketch the $x$-$y$ plane on which the random point $(X,Y)$ lies. Then, for any real number $z$, $X+Y \leq z$ if and only if $(X,Y)$ lies in the half-plane on or below the straight line $x+y = z$ (which passes through the points $(0,z)$ and $(z,0)$. Thus, $F_Z(z) = P\{X+Y\leq z\}$ is just the integral of the joint pdf over this region. For a fixed $y$, $x$ can be anything from $-\infty$ to $z-y$ and it will still be true that $x+y \leq z$, etc. etc. etc. On your second question, reflect that if $\min\{X,Y\} = w$ for some $w, 0<w<0.5$, then $X+Y$ must be at least $2w$, no? $\endgroup$ – Dilip Sarwate Jan 18 '14 at 15:30
  • $\begingroup$ Thanks. I just wanted to express everything formally and didn't found the way by doing this. Also, how do you differentiate with respecto to $z$ without working out the integrals? $\endgroup$ – V. Galerkin Jan 18 '14 at 16:02
  • $\begingroup$ See my comment on a previous answer of mine that also used the concept of differentiating an integral. $\endgroup$ – Dilip Sarwate Jan 18 '14 at 16:06

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