Find the MARGINAL survival function

Question

Find the marginal survival of $X$ when $$ S(x,y) = (1-x)(1-y)(1+\frac{xy}{2}),0<x<1,0<y<1$$

So if we have a joint pdf $f(x,y)$, then the marginal is $f(x) = \int_{-\infty}^\infty f(x,y)dy. $.so would the same logic be applied here? If so I get:

$$f(x) = \int_0^1 S(x,y) dy= ...=(1-x)(\frac{x}{12}+\frac{1}{2})$$.

Then $$ S(x) = \int_x^1 f(x) dx = \int_x^1(1-x)(\frac{x}{12}+\frac{1}{2})$$

Is this right?

Answer 1

A simpler way is to evaluate the joint survival at either $x=0$ or $y=0$.

The joint survival function is:

$$ S(x,y)=\mathbb P(X>x,Y>y) $$

and the marginals survival functions are $$ S_X(x) = \mathbb P(X>x) \\ S_Y(y) = \mathbb P(Y>y) $$

since $X$ and $Y$ belong to $(0,1)$ then \begin{align*} \mathbb P(X>x) &= \mathbb P(X>x, Y>0) \\ &= S(x,0) \\ &= 1-x \end{align*}

The same apply to $Y$, $P(Y>y) = S(0,y) = 1-y$.

Thus $X$ and $Y$ are uniformly distributed over $(0,1)$