Let's say Z=X+Y, where X and Y are independent uniform random variables with range [0,1]. Then the PDF is&nbsp;f(z)={zfor&nbsp;0&lt;z&lt;12−zx&gt;for&nbsp;1≤z&lt;20otherwise.How was this PDF obtained?

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Let&#039;s say Z=X+Y, where X and Y are independent uniform random variables with range [0,1]. Then the PDF is&amp;nbsp;f(z)={zfor&amp;nbsp;0&amp;lt;z&amp;lt;12−zx&amp;gt;for&amp;nbsp;1≤z&amp;lt;20otherwise.How was this PDF obtained?

sonSnubsreose6v · Accepted Answer

If we want to use a convolution, let fX be the full  density function of X, and let fY be the full density function of  Y. Let Z=X+Y. Then   fz(z)=∫−∞∞fX(x)fY(z−x)dx.  Now let us apply this general formula to our particular case. We will have fZ(z)=0forz&amp;lt;0, and also for z≥2. Now we deal with the interval from 0 to 2. It is useful to break this down into two cases (i) 0&amp;lt;z≤1and(ii)1&amp;lt;z&amp;lt;2.  (i) The product fX(x)fY(z−x)is 1 in some places, and 0 elsewhere. We want to make sure we avoid calling it 1 when it is 0. In order to have fY(z−x)=1,we≠edz−x≥0,ti^s,x≤z. So for (i), we will be integrating from x = 0 to x = z. And easily  ∫0z1dx=z  Thus fZ(z)=zfor0&amp;lt;z≤1  (ii) Suppose that 1&amp;lt;z&amp;lt;2. In order to have fY(z−x) to be 1, we need z−x≤1,ti^s,we≠edx≥z−1. So for (ii) we integrate from z−1 to 1. And easily  ∫1z−11dx=2−z  Thus fZ(z)=2−zfor1&amp;lt;z&amp;lt;2.   Another way: (Sketch) We can go after the cdf FZ(z)ofZ, and then differentiate. So we need to find Pr(Z≤z).  For a few fixed z values, draw the lines with equation x+y=z on an x-y axis plot. Draw the square S with corners (0,0), (1,0), (1,1), and (0,1).   Then Pr(Z≤z) is the area of the part S that is &quot;below&quot; the line x+y=z.  That area can be calculated using basic geometry. For example, when z  is 2, the whole square area is under the line so Pr=1. There is a switch  in basic shape at z=1.

ol3i4c5s4hr · Accepted Answer

By the hint of jay-sun, consider this idea, if and only if fX(z−y)=1when0≤z−y≤1. So we get   z−1≤y≤z  however, z∈[0,2], the range of y may not be in the range of [0,1] in order to get fX(z−y)=1, and the value 1 is a good splitting point. Because z−1∈[−1,1].  Consider (i)ifz−1≤0then−1≤z−1≤0ti^sz∈[0,1], we get the range of y∈[0,z]sin⁡cez∈[0,1]. And we get  ∫−∞∞fX(z−y)dy=∫0z1dy=zifz∈[0,1].  Consider (ii)ifz−1≥0ti^sz∈[1,2], so we get the range of y ∈[z−1,1],and∫−∞∞fX(z−y)dy=∫z−111dy=2−zifz∈[1,2]  To sum up, consider to clip the range in order to get fX(z−y)=1.

Let's say Z=X+Y, f(z)=z for 0<z<1 2-z x> for 1<=z<2 0 otherwise

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