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I am having trouble understanding the solution to this problem. The problem define $X_1$ and $X_2$ to have the joint pdf $f(x_1, x_2) = 15x_1^2 x_2$, $0 < x_1 < x_2 < 1$, zero elsewhere. It then asks to calculate the marginal pdfs and $P(X_1 + X_2 \leq 1)$. I was able to get the marginal pdfs for $X_1$ and $X_2$, but was a bit confused on the solution of $P(X_1 + X_2) \leq 1$. They set up the integral as:

$$ P(X_1 + X_2 \leq 1) = \int_0^{\frac{1}{2}} \int_{x_1}^{1 - x_1} f(x_1, x_2) dx_2 dx_1 $$

And arrived at the answer of $\frac{5}{64}$. I have tried drawing the line and understood why they set up the integral like that, but I don't understand why doing either of the following ways do not give me the same answer.

  1. Does the order of integration matter? I was trying with this integral and get a different result.

$$ P(X_1 + X_2 \leq 1) = \int_{x_1}^{1 - x_1} \int_0^{\frac{1}{2}} f(x_1, x_2) dx_1 dx_2 $$

  1. Also, why doesn't this work?

$$ P(X_1 + X_2 \leq 1) = \int_{\frac{1}{2}}^1 \int_{x_2}^{1 - x_2} f(x_1, x_2) dx_1 dx_2 $$

Edit: I now understand why the second way does not make sense, but still stuck on the first way.

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    $\begingroup$ I suggest you draw a picture. draw a unit square and the line $y+x=1$. Look at the area below this line and inside the square... $\endgroup$
    – utobi
    Oct 8, 2022 at 18:55
  • $\begingroup$ Thanks! I actually have drawn it and understood why they did it like it, but still don't understand why if I do like the other 2 ways I showed, I won't arrive at the same answer. $\endgroup$
    – DiMario
    Oct 8, 2022 at 19:10
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    $\begingroup$ Re order of integration: see en.wikipedia.org/wiki/Fubini%27s_theorem. $\endgroup$
    – whuber
    Oct 8, 2022 at 20:02

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You already answered your own question for 2. With regards to 1., the order of integration does not matter, and it should generate the same answer. Whuber posted a link in the comments explaining how Fubini's Theorem shows that.

Also, the math is below to show how the double integral, when swapped, results in the same answer. $$ Original: \int_{x_{1}}^{1-x_{1}}\int_{0}^{\frac{1}{2}} 15{x_{1}}^{2}x_{2} dx_{1} dx_{2} $$

Integrate with respect to x1 $$ \int_{x_{1}}^{1-x_{1}}\frac{15{\frac{1}{2}}^{3}x_{2}}{3} dx_{2}$$

Simplify and then substitute in the boundaries used to evaluate dx1 within the x2 boundaries for x1 $$ \int_{0}^{\frac{1}{2}} \frac{15}{24}x_{2} dx_{2} $$

Then integrate with respect to x2 and you will see you'll get the same result $$ \frac{15}{24*2}\frac{1}{2}^{2} = \frac{15}{192} = \frac{5}{64} $$

I also think it'd be helpful to others if you explain how you figured out part 2.

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