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?

  • 1
    $\begingroup$ The "marginal survival" is the survival function of the marginal distribution. Seems to be self-study? $\endgroup$
    – Yves
    Oct 19, 2020 at 8:45

1 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 $(0,1)$

  • $\begingroup$ Thanks! So would the cumulative incidence function of $T_1$ simply be $F(X) = 1 - S(X)$? $\endgroup$
    – CCZ23
    Oct 20, 2020 at 9:11
  • $\begingroup$ @CCZ23 The cumulative incidence function is $h(x) = \mathbb P(X \leq x , X \leq Y)$, which is different from $F(x) = \mathbb P(X \leq x)$. In order to find $h(x)$ it you can compute the joint density $f(x,y)$ and integrate it over the appropriate set. Detailing it would require more than a comment but I found $$h(x) = \frac{12x -9x^2 +6x^3-3x^4}{12}$$ If you struggle to find $h(x)$ maybe you can ask another question specifically for this. $\endgroup$
    – periwinkle
    Oct 20, 2020 at 20:48

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