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Consider two independent, identically distributed random vectors $\vec{X}=(X_1, X_2, ...,X_N)$ and $\vec{Y}=(Y_1, Y_2, ...,Y_N)$, where $X_i \sim U[0,1]$ and $Y_i \sim U[0,1]$.

What is the distribution of the rv $Z$ obtained from the scalar product between $\vec{X}$ and $\vec{Y}$ (that is

$Z = <\vec{X},\vec{Y}> = \sum\limits_{i=1}^{N}X_iY_i $)?

So far, my approach to solve this is to start from $Z|\vec{Y}$ which is the sum of $N$ uniform distributions $U[0,Y_i]$. While the sum of $N$ non-identically distributed uniforms is known, with this method I cannot get a general result for any value of $N$. I would like to know if there is a simpler approach, or even better, if this problem has already been solved (I know similar problems, e.g. when the vectors are on the unit, $n-$sphere or when the vectors are gaussian).

vaz
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  • By "iid vectors" do you mean that $X_1, X_2, ...,X_N$ are independent and identically distributed as $U[0,1]$? Similarly $Y_1, Y_2, ...,Y_N$ are iid $U[0,1]$? – rookie Mar 16 '17 at 10:53
  • Exactly. Should I rephrase the question? – vaz Mar 16 '17 at 11:24
  • Just include the independence phrase in the first statement of the details. – rookie Mar 16 '17 at 15:54
  • If one can find the density of $XY$ then the density of $Z$ will be $N$ times convolution of the density of $XY$. Agree? – rookie Mar 16 '17 at 15:57
  • Yes, but I'm not sure if that is a simpler approach. – vaz Mar 17 '17 at 08:12
  • If $N$ times convolution is hard then find the characteristic function (https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)) of $XY$, and then characteristic function of $Z$ will be just the $N^\text{th}$ power of $XY$'s characteristic function. Fourier transform is used to go between the density and the characteristic function of a random variable. – rookie Mar 17 '17 at 09:35
  • I will dig into that as soon as I find the time, and post my answer (if I found) it. Thanks for the advice. – vaz Mar 21 '17 at 08:21
  • So the density of $Z=XY$ is $f_Z(z)=-\log(z)$, as shown in this post http://math.stackexchange.com/questions/659254/product-distribution-of-two-uniform-distribution-what-about-3-or-more The characteristic function is $E[\exp(j\omega Z)]=\int -\log(z)\exp[j\omega z]dz$, which doesn't look easy to solve... – vaz Mar 28 '17 at 08:49

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