# Find the MARGINAL survival function

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

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?

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

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 $$(0,1)$$

• Thanks! So would the cumulative incidence function of $T_1$ simply be $F(X) = 1 - S(X)$? Oct 20, 2020 at 9:11
• @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. Oct 20, 2020 at 20:48