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Let X, Y be independent random variables each with the uniform distribution on the interval [0, 1].

Compute the cumulative density function of $W = X + Y$

I know that when $1<w<2$, we have to compute the trapezoid with the given vertices, but in fact shouldn’t a trapezoid have 4 vertices instead of 5 $(0, 0)$;$(1, 0)$;$(0,1)$;$(w-1,1)$ and $(1, w-1)$ and I thought then it might be calculated as 1- the area of the triangle above $x+y$ but from where did we get the $(w-2)^2$?

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  • $\begingroup$ I believe that you mean the cumulative density function, since the probability density function is clear from the definition of $W$ and the distributions of $X$ and $Y$. (And in fact, is the first thing shown in your image!) $\endgroup$ – rwbogl Jul 23 '18 at 0:37
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    $\begingroup$ Yes I mean in the cumulative density function $\endgroup$ – Roy Rizk Jul 23 '18 at 0:38
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    $\begingroup$ Discussion and plots in my answer are intended to help you visualize the steps in the analytical solution. // Meanwhile, @rwbogl (+1) gave a nice start on the details of the analytical solution, which I hope your will Accept. Try plotting the formulas for the CDF and PDF when you get them. // Also, perhaps edit a little of your own work into your Question to keep it from being 'Closed'. $\endgroup$ – BruceET Jul 23 '18 at 1:22
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You are right that the shape described in the text is not a trapezoid. It is an irregular pentagon. You can see it in the lower left portion of the square in the figure below.

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As you said, the area of this pentagon is the area of the square (which is $1$) minus the area of the triangle in the upper right corner. The legs of the triangle have length $2 - w$ because $$ 1 - (w - 1) = 2 - w. $$

Also notice that if $w = 2,$ the only possible $x,y$ value is $(1,1),$ which is consistent with shrinking the triangle down to a single point (leg length $0$), whereas if $w = 1$ then we could have $(x,y) = (0,1)$ or $(x,y) = (1,0)$ or anything on the segment between those points--in other words, a triangle with legs of length $1$ covering half of the square. For $1 < w < 2$ we get something in between, like the triangle shown in the figure.

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Let's trace out what your image says.

Set $W = X + Y$, where $X, Y \sim \text{Uniform}(0, 1)$. We want to find the c.d.f. of $W$; call this $F_W$.

By definition, $$F_W(w) = P(W \leq w) = \int_{R_w} dx dy,$$ where $R_w$ is the intersection of $\{(x, y) \mid x + y \leq w\}$ and $[0, 1]^2$. That is, everything in the unit square with $x + y \leq w$.

As mentioned, the nature of this region changes for different $w$. If we draw a picture (draw it!), we can see that the region is a triangle for $0 \leq w \leq 1$, and a trapezoid when $1 < w < 2$. Further, if we label our picture a little (label it!), we can see that the vertices are what the image says.

To integrate this trapezoidal region, we need two integrals: $$\int_{R_w} dx dy = \int_0^{w - 1} \int_0^1 dy dx + \int_{w - 1}^1 \int_0^{w - x} dy dx.$$

Evaluating these gives $- \frac{w^{2}}{2} + 2 w - 1$, which is equal to the given expression for the case $1 < w < 2$.

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Quoting from the posted information for finding $P(W \le w):$ When $0 < w \le 1,$ the region $R_w$ is a triangle with vertices $(0,0), (0, w),$ and $(w, 0).$ Then this triangle has area $w^2/2.$ Because the distribution of $X$ and $Y$ is uniform on the square with vertices $(0,1)$ and $(1,1),$ probabilities are proportional to areas. Thus $P(W \le w) = w^2/2,$ for $0 < w \le 1.$

If you want the density function, then differentiate the CDF, first for $0 < w < 1$ and then for $1 <w \le 2.$


A simulation of 50,000 values of $W$ makes it easy to show some areas, the PDF and the CDF. In the figure below, the first plot shows (red) the area corresponding to $P(X < 0.5);$ the second shows (green) the area corresponding to $P(W < 1.6);$ the third shows the empirical CDF, which suggests the shape of the CDF of $W$ you are asked to compute; and the fourth shows a histogram of simulated values of $W.$ which suggests the triangular shape of the density function of $W.$

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The empirical CDF (ECDF) jumps up by $1/50,000$ at each observed value of $W.$ Usually, an ECDF gives a more precise indication of the CDF than a histogram gives of the PDF (because information is lost when data are put into histogram bins).

Addendum: R code for figure per Comments.

par(mfrow=c(2,2))  # enables 4 panels per plot
m = 50000;  x = runif(m);  y = runif(m);  w = x+y
 plot(x, y, pch=".", main="Joint Dist'n of X and Y with P(W < .5)")
  points(x[w < .5], y[w < .5], pch=".", col="red")  # read [] as 'such that'
 plot(x, y, pch=".", main="Joint Dist'n of X and Y with P(W < 1.6)")
  points(x[w < 1.6], y[w < 1.6], pch=".", col="green2")
 plot(ecdf(w), main="Emplrical CDF of Observed Values of W")  # 'ecdf' procedure
 hist(w, prob=T, col="skyblue2", main="Histogram of Observed Values of W")
par(mfrow=c(1,1))

Notes: points puts points onto an existing plot. ecdf includes data sort and and cumulative tally, then returns x and y values ready for plotting (actually a stair-step plot but seems a curve with 50,000 points at this resolution); argument prob=T of hist function invokes vertical density scale.

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  • $\begingroup$ This is an amazing empirical answer for this question. How are you estimating the c.d.f. numerically? $\endgroup$ – rwbogl Jul 23 '18 at 1:27
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    $\begingroup$ Thanks, easy in R. I'll clean up the code and post it as an addendum $\endgroup$ – BruceET Jul 23 '18 at 1:29

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