Multivariate Distribution with $x+y>1$ constraint

Question

If the joint probability density of $X$ and $Y$ is given by

$$f(x,y)=\left\{\begin{matrix} 24xy,0<x<1,0<y<1,x+y<1 & & & \\ 0, elsewhere& & & \end{matrix}\right.$$

Find $P(X+Y<\frac{1}{2})$

I don't understand how the given answer comes from. How do I determine the lower limit and the upper limit for the double integral? Thanks in advance.

Answer 1

$$\begin{aligned} P(X+Y<\frac{1}{2}) &= \iint_{\Bbb{R}^2} f(x,y) \cdot 1_{0 < x+y < 1/2} \, dxdy \\ &= \int_0^{1/2} \int_0^{1/2} 24xy \cdot 1_{0 < x+y < 1/2} \, dydx \\ &= \int_0^{1/2} \int_{0}^{1/2-x} 24xy dy dx \\ &= \int_0^{1/2} 12x (\frac12 - x)^2 dx \\ &= \int_0^{1/2} 12x (\frac14-x+x^2) dx \\ &= \int_0^{1/2} (3x-12x^2+12x^3) dx \\ &= \frac32(\frac12)^2 -4(\frac12)^3 + 3(\frac12)^4\\ &= \frac{1}{16} \end{aligned}$$

Answer 2

The probability $\mathbf{P}(X + Y < \frac12)$ is equivalent to the expectation $$ \mathbf{E}[\mathbf{1}_A] = \int_A f(x,y) dx dy$$ where $A = \{0 < x < 1, \, 0 < y < 1, \, 0 < x + y < \frac12 \}$.

Therefore we can solve this using standard double integral formulations, by first integrating out over $x$, and then $y$. If we allow $x$ to range between $0$ and $1$, then for any given $x$ we are constrained to have $$0 < y < \frac12 - x$$ else we violate the condition that $(x,y) \in A$. So this inequality above gives us the bounds of integration of $y$. So that is

\begin{align*} \mathbf{P}\left(X + Y < \frac12\right) &= \int_A f(x,y) dx dy \\ & = \int_0^1 \int_{0}^{\frac12 - x} f(x,y) d yd x \\ & = 24 \int_0^1 x \left( \int_0^{\frac12 - x} y dy\right) dx, \end{align*} from here you can solve this using standard integration formulae for polynomials.

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If you are happier with the notions of conditional probablity, than you are with double integrals, then one can also reach this point relying purely on probabilistic notions.

We note that by the law of total expectation \begin{align*} \mathbf{E}[\mathbf{1}_A] &= \mathbf{E} \big[ \mathbf{E}[\mathbf{1}_A |X] \big] \\ &= \mathbf{E} \big[ \mathbf{P}[Y < \textstyle{\frac12} -X |X] \big] \\ &= \int_{0}^1 f_X(x) \mathbf{P}[Y < \textstyle{\frac12} -X |X = x] d x \\ &= \int_{0}^1 f_X(x) \mathbf{P}[Y < \textstyle{\frac12} -x |X = x] d x \\ &= \int_{0}^1 f_X(x) \left( \int_0^{\frac12 - x} f_{Y|X}(y|x)d y\right) d x \\ &= \int_{0}^1 \int_0^{\frac12 - x} f_X(x) f_{Y|X}(y|x)d y d x \\ &= \int_{0}^1 \int_0^{\frac12 - x} f_{X,Y}(x,y)d y d x \\ \end{align*} where in the final line we used the identity that $f_{Y|X}(y|x) = f_{X,Y}(x,y)/f_X(x)$.