I am stuck with this tutorial question in one of my stats module and I would greatly appreciate some help:

Let $X1$ and $X2$ be independent random variables with $a = 0$ and $b = 1$ i.e. $X1$ and $X2$ are uniformly distributed over 0 to 1.

How do you find the distribution function of $Y = X1 + X2$ for $1<y<2$ i.e. what is $F(y)$?


2 Answers 2



We can first derive the PDF using convolution of two PDFs:

Case 1: If $0 \leq y \leq 1$, then $f_{X_1}(y - x_2) = 1$ if $ 0\leq x_2 \leq y$, and $f_{X_1}(y-x_2) = 0$ if $x_2 > y$. This means that \begin{equation} \int_0^1 f_{X_1}(y-x_2) dx_2 = \int_0^y 1 dx_2 = y \end{equation}

Case 2: If $1 < y < 2$, then $f_{X_1}(y-x_2) = 1$ if $y-1\leq x_2 \leq 1$, and $f_{X_1}(y-x_2) = 0$ otherwise. So \begin{equation} \int_0^1 f_{X_1}(y-x_2) d x_2 = \int_{y-1}^1 1 dx_2 = 2-y \end{equation}

Conclusion: $f_Y(y) = y$ if $0 \leq y \leq 1$ and $2-y$ if $1 \leq y \leq 2$. Otherwise, it is zero.


This means the CDF, which is defined as follows \begin{equation} F_Y(y) = \int_{-\infty}^y f_Y(y) dy \end{equation}

Case 1: If $y < 0$: Clearly $F_Y(y) = 0$

Case 2: If $0 <y < 1$, then \begin{equation} F_Y(y) = \int_{0}^y f_Y(t) dt = \int_0^y t dt = \frac{y^2}{2} \end{equation}

Case 3: If $1<y<2$, then \begin{equation} F_Y(y) = \int_{0}^y f_Y(t) dt = \int_0^1 t dt + \int_1^y 2-t dt =\frac{1}{2} -\dfrac{y^2-4y+3}{2} \end{equation}

Case 4: If $y > 2$, then $F_Y(y) = 1$.


We need this probability: $F_Y(y)=P(Y\leq y)=P(X_1+X_2 \leq y)$ Consider a unit square in $[0,1]\ \text{x} \ [0,1]$. The Joint PDF .is distributed uniformly on this square, with value $1$. We draw $x_1+x_2=y$ line and integrate the Joint PDF below the line, which is equivalent to finding the area of the square below the line.

There are two cases: $0<y<1$ and $2>y>1$. I guess you need only $y>1$ part, which is "For you to find :)" I believe. You just need to draw the square, draw the line, find the area of the square above the line and subtract from $1$.

Note: There is also a solution with convolution if you're interested in.

  • $\begingroup$ Thank you for the guide! I could find 0 < y < 1 which is 0.5(y^2), but I seem to get stuck moving to 1 < y < 2. Do you mind sharing the integrals for the area of the square above the line and i try to figure out the reasoning myself? @gunes $\endgroup$ Sep 25, 2018 at 13:18
  • $\begingroup$ Writing the integral is harder here, since the joint PDF is uniform, it is much easier to find the area of the triangle above the line and then subtract it . But, the integral of $P(Y\geq y)$ is $$\int_{y-1}^{1}{\int_{x_1}^{1}{1 \ dx_2dx_1}}$$. $\endgroup$
    – gunes
    Sep 25, 2018 at 13:55

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