Is sum of two uniform random variables is uniformly distributed?

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

• Is the sum of two dice uniformly distributed? – URL Jan 19 at 7:25

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.

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 $$0 and $$0 and zero otherwise. There are three cases to consider:

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

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

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

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

• (assuming independence) – Teepeemm Jan 19 at 15:17
• @Teepeemm Yes, that is the general assumption when no information is given about the joint distribution of random variables. – Math1000 Jan 19 at 16:22
• 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 $y$? – student Jan 20 at 14:03
• Yes. But be careful as to whether $y\in (0,t)$, $y\in(t,2t)$, or $y\in(2t,3t)$. – Math1000 Jan 21 at 9:12