Consider two independent random variables $X$ and $Y$. Let $$f_X(x) = \begin{cases} 1 − x/2, & \text{if $0\le x\le 2$} \\ 0, & \text{otherwise} \end{cases}$$.Let $$f_Y(y) = \begin{cases} 2-2y, & \text{if $0\le y\le 1$} \\ 0, & \text{otherwise} \end{cases}$$. Find the probability density function of $X + Y$.

Can anyone show me a step by step solution to this problem?

I've been applying this theorem to solve the problem with limited success enter image description here

  • $\begingroup$ It is the convolution of the two densities, assuming independence. $\endgroup$ – Gary. Jul 10 '15 at 4:10

Let $Z=X+Y$. We know $0\leq Z\leq 3$ but for the density of $Z$ we have three cases to consider due to the ranges of $X$ and $Y$.

If $0\leq z\leq 1$:

\begin{eqnarray*} f_Z(z) &=& \int_{x=0}^{z} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=0}^{z} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=0}^{z} \\ &=& 2z - \dfrac{3}{2}z^2 + \dfrac{1}{6}z^3. \end{eqnarray*}

If $1\lt z\leq 2$:

\begin{eqnarray*} f_Z(z) &=& \int_{x=z-1}^{z} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=z-1}^{z} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=z-1}^{z} \\ &=& \dfrac{7}{6} - \dfrac{1}{2}z. \end{eqnarray*}

If $2\lt z\leq 3$:

\begin{eqnarray*} f_Z(z) &=& \int_{x=z-1}^{2} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=z-1}^{2} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=z-1}^{2} \\ &=& \dfrac{9}{2} - \dfrac{9}{2}z + \dfrac{3}{2}z^2 - \dfrac{1}{6}z^3. \end{eqnarray*}

  • $\begingroup$ Mick, I'm not completely sure how you chose the bounds for the different integrals- do you mind briefly explaining? $\endgroup$ – aman1230 Jul 10 '15 at 16:56
  • $\begingroup$ @aman1230 For $0\leq z\leq 1:\;$ Here, $x$ must be less than $z$ since $y$ can't be negative. So we must have $0\lt x\lt 1$. For $1\lt z\leq 2:\;$ Again, $x$ must be less than $z$ since $y$ can't be negative. Also, since $y\lt 1$, we must have $x$ at least $z-1$. Thus, $z-1\lt x\lt z$. For $2\lt z\leq 3:\;$ Again, since $y\lt 1$, we must have $x$ at least $z-1$. Also, $x$ cannot exceed $2$ so that instead of $z$ is the upper limit here. Thus, $z-1\lt x\lt 2$. I hope that helps explain it. $\endgroup$ – Mick A Jul 10 '15 at 17:25
  • $\begingroup$ This is quite intuitive Mick. I appreciate your help. Thanks. $\endgroup$ – aman1230 Jul 11 '15 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.