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Let $X$ and $Y$ be independent random variables taking values in $[0,1]$ where $X$ is uniform. Question is, what distribution on $Y$ will yield a uniform distribution on $[0,2]$ for the sum $Z=X+Y$?

Somehow, by inspection, since the distribution of the sum is $F_Z(z)=\int F_Y(z-x)\,dF_X(x)$, I figured out $Y=0$ or $1$ with prob. $1/2$ each works.

Is there an intutive way to guess the answer?

Ashok
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2 Answers2

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Intuitively $Y$ must be symmetric if $X$ and $Z$ are.

Intuitively $Y$ cannot take values greater than $1$ or less than $0$ as $Z$ could then be outside $[0,2]$.

Intuitively $Y$ cannot take values in the open interval $(0,1)$ as if it did at or near a value $y$ then the density of $Z$ in $(y,y+\delta y)$ would be greater than the density of $Z$ in $(0,\delta y)$.

So your result is the intuitive answer, and is easily extended to $Z$ uniform in $[k,k+n]$ for integer $n$, and shows why $Z$ cannot be uniform over an interval which is not of integer length.

Henry
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  • Thanks for the nice intuition. However, could you elaborate a little on the 3rd intuition. I couldn't get how putting mass at the end points doesn't affect whereas in between does affect. – Ashok May 01 '13 at 12:53
  • @Ashok: If you had a point mass at $0$ and a similar point mass at $\frac12$ then the density would be twice as high just above above $\frac12$ as just above $0$ – Henry May 07 '13 at 21:35
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Note that the moment generating function of such a $Y$ if it exists is $\text{E}(\exp t Z)/\text{E}( \exp tX) = \frac{1}{2}\frac{\exp 2t - 1}{\exp t - 1} = \frac{1}{2}\left( \exp t + 1 \right)$ which is the moment generating function of a random variable having equal point masses on $\{0,1\}$.

Arin Chaudhuri
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