Let $X_1 \sim Uniform(0,1)$, and $X_2 \sim Uniform(0, x_1)$, where $x_1$ is the realized value of $X_1$. Find $P(X_1 + X_2 \geq 1)$.

I know that I need the joint distribution of $X_1$ and $X_2$.

$f_1(x_1) = 1, \ 0 < x_1 < 1$

$f_{2|1}(x_2|x_1) = \frac{1}{x_1}, \ 0 < x_2 < x_1$

$f_{1,2}(x_1, x_2) = f_{2|1}(x_2|x_1)f_1(x_1) = \frac{1}{x_1}, \ 0 < x_2 < x_1 < 1$.

What are the limits of integration when integrating this joint PDF to get the desired probability?

  • $\begingroup$ What relation do $X_1$ and $X_2$ have? Independent? $\endgroup$
    – user158565
    Commented Nov 30, 2018 at 1:51
  • $\begingroup$ @user158565 They can't be independent; the form of dependence is given in the question. $\endgroup$
    – Glen_b
    Commented Nov 30, 2018 at 5:01
  • $\begingroup$ @Glen_b Answer was modified. $\endgroup$
    – user158565
    Commented Nov 30, 2018 at 5:19

2 Answers 2


enter image description here

Here is some hint:

You already get the joint pdf, and the range of $X_1$, and $X_2$ as showed in graph black and red area.

The probability of the event = integral of pdf on the area that event defined. In your case, $X_1+X_2>1$ is the red area in the graph above. So $$\Pr(X_1+X_2>1) = \int_?^?\int_?^? \frac 1{x_1}dx_1dx_2$$ So you need find the limits of the integrals, then you get the answer.

  • $\begingroup$ Thanks, that's really helpful. I think that it's easier to integrate with respect to $x_2$ first, from $1- x_1$ to $x_1$. Then, I integrate with respect to $x_1$, from $0.5$ to $1$. $\endgroup$
    – MSE
    Commented Dec 2, 2018 at 14:07
  • $\begingroup$ You are right. Graph is very helpful for this kind of questions. $\endgroup$
    – user158565
    Commented Dec 2, 2018 at 16:40

My advice with figuring out limits is draw a picture (by hand will generally suffice, though):

plot of X2 vs X1 with joint density indicated by dot density, X1 + X2 = 1 line marked in

You should be able to figure out the upper and lower limits from that. Consider which variable to integrate over first; one may be a tad easier than the other.

There's also a symmetry you can exploit which simplifies the slightly trickier of the two integrals, which may save a line.


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