# If $X_1$ and $X_2$ are uniformly distributed random variables with parameters $0$ and $1$, what is the distribution of $Y = X_1 + X_2$?

I was doing a recap of the probability theory I had last year and even though this question shouldn't be hard, it is somehow confusing me immensly.

Clearly, if we have $$X_1, X_2$$ belonging to $$U[0, 1]$$, then $$Y = X_1 + X_2$$ must have a support of $$Y$$ belonging to $$[0, 2]$$, as both $$X_1$$ and $$X_2$$ belong to $$[0,1]$$. By definition, the pdf of $$U[0, 1]$$ is simply $$1$$. Now, if I try to find the cdf of $$Y$$, I find that:

$$F_Y(y) = \int_0^y\int_0^{y-x2}f(x_1, x_2)dx_1dx_2 = \int_0^y\int_0^{y-x2}1dx_1dx_2 = \int_0^y(y-x_2)dx_2 = y^2/2$$.

For a distribution function, if $$y$$ tends to its limit, then $$F_Y(y)$$ tends to $$1$$. However, if I let $$y = 2$$, then $$F_Y(y) = 2$$ as well, which can't be.

I must be doing something incredibly stupidly wrong. Please tell me what it is.

You've got (in your outer integral) $$x_2$$ going from $$0$$ up to $$y$$, which can't be right, as $$x_2$$ never gets bigger than $$1$$.

You haven't mentioned whether or not $$X_1$$ and $$X_2$$ are independent, but if that is the case then you should be able to break the integral down to two cases,

case 1) $$0 < y < 1$$

case 2) $$1 < y < 2$$

and be more careful about the limits of integration you use. Remember, $$f(x_1, x_2) = 0$$ outside of $$[0,1] \times[0,1]$$.

(If you can't assume that $$X_1$$ and $$X_2$$ are independent then you probably won't be able to get any nice closed-form expression for $$F_Y$$.)

• Thank you so much. – RJAL Feb 22 at 22:27
• Glad to help, @RJAL. If this response answers your question you can accept it by clicking on the check mark on the left side. – JonathanZ Feb 22 at 22:48
• The problem statement doe not say $X_1$ and $X_2$ are independent, although it is needed for the definition of $f(x_1,x_2)$. – herb steinberg Feb 22 at 22:55
• @herbsteinberg: Excellent point. From everything else the OP wrote I had assumed that was the case, but they hadn't said so. I'll update to include that. – JonathanZ Feb 22 at 23:30