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For example, $X_1, X_2 \sim U[0,t]$. Does it imply that $2X_1+X_2 \sim U[0,3t]$?

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    $\begingroup$ Is the sum of two dice uniformly distributed? $\endgroup$ – URL Jan 19 at 7:25
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If $X_2$ is always equal to $X_1,$ and $X_1\sim\operatorname{Uniform}[0,t],$ then $2X_1+X_2\sim\operatorname{Uniform}[0,3t].$ At the opposite extreme, if $X_1,X_2\sim\operatorname{Uniform}[0,t]$ and $X_1,X_2$ are independent, then $2X_1+X_2$ is not uniformly distributed. You haven't told use the joint distribution of $X_1,X_2,$ but only how each one separately is distributed.

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A linear combination of uniform random variables is in general not uniformly distributed. In this case we have $Y=2X_1+X_2$, where $2X_1\sim\mathrm{Unif}(0,2t)$ and $X_2\sim\mathrm{Unif}(0,t)$. The density of $2X_1$ is $f_{2X_1}(x) =\frac1{2t}\mathsf 1_{(0,2t)}(x)$, and the density of $X_2$ is $f_{X_2}(x) = \frac 1t\mathsf 1_{(0,t)}(x)$. Therefore the density of the sum is given by convolution: $$ f_Y(y) = f_{2X_1}\star f_{X_2}(y) = \int_{\mathbb R} f_{2X_1}(x)f_{X_2}(y-x)\ \mathsf dx = \int_{\mathbb R}\frac1{2t}\mathsf 1_{(0,2t)}(x)\cdot\frac 1t\mathsf 1_{(0,t)}(y-x)\ \mathsf dx. $$ Now, $\mathsf 1_{(0,2t)}(x)\mathsf 1_{(0,t)}(y-x)$ is equal to one when $0<x<2t$ and $0<y-x<t$ and zero otherwise. There are three cases to consider:

If $0<y<t$ then the convolution integral is $$ \int_0^y \frac1{2t^2}\ \mathsf dx = \frac y{2t^2}. $$

If $t<y<2t$ then the convolution integral is $$ \int_{y-t}^y \frac1{2t^2}\ \mathsf dx = \frac1{2t}. $$

If $2t<y<3t$ then the convolution integral is $$ \int_{y-t}^{2t} \frac1{2t^2}\ \mathsf dx = \frac{3 t-y}{2 t^2}. $$

Hence the density of $Y$ is given by $$ \frac y{2t^2}\mathsf 1_{(0,t)}(y) + \frac1{2t}\mathsf 1_{(t,2t)}(y) + \frac{3 t-y}{2 t^2}\mathsf 1_{(2t,3t)}(y). $$

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  • $\begingroup$ (assuming independence) $\endgroup$ – Teepeemm Jan 19 at 15:17
  • $\begingroup$ @Teepeemm Yes, that is the general assumption when no information is given about the joint distribution of random variables. $\endgroup$ – Math1000 Jan 19 at 16:22
  • $\begingroup$ If I want to find $P(Y<c)$, I should integrate $$\frac y{2t^2}\mathsf 1_{(0,t)}(y) + \frac1{2t}\mathsf 1_{(t,2t)}(y) + \frac{3 t-y}{2 t^2}\mathsf 1_{(2t,3t)}(y). $$ with respect to $y$? $\endgroup$ – student Jan 20 at 14:03
  • $\begingroup$ Yes. But be careful as to whether $y\in (0,t)$, $y\in(t,2t)$, or $y\in(2t,3t)$. $\endgroup$ – Math1000 Jan 21 at 9:12

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