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I'm trying to solve a problem with various steps. There's the following joint pdf

$f_{XY}(x,y)= \begin{cases} x+y, &\text{if }0<x<1 \text{ and } 0<y<1\\0, &\text{otherwise} \end{cases} $

and one of the questions states that you need to find $P(Y<0.6|X>0.9)$. Given that I need the conditional density function, first I got the marginal distribution of $X$ which is $f_X(x)=x+\dfrac 12$, and then the conditional distribution would be $f_{Y|X=x}(y)=\dfrac{x+y}{x+\dfrac 12}=\dfrac {2(x+y)}{2x+1}$.

My question is the following: in order to obtain $P(Y<0.6|X>0.9)$, what do I need to do? Since X is not equal but rather larger than $0.9$, do I calculate the conditional distribution with the result of integrating $f_X$ from $0.9$ to $1$, and then divide $(x+y)$ by that result? All examples I've found X asume X takes a fixed value.

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My question is the following: in order to obtain $P(Y<0.6|X>0.9)$, what do I need to do?

Directly use the definition for conditional probability : $$\mathsf P(Y<0.6\mid X>0.9)~{=\dfrac{\mathsf P(Y<0.6\cap X>0.9)}{\mathsf P(X>0.9)}\\=\dfrac{\int_0^{0.6}\int_{0.9}^1 (x+y)~\mathsf d x~\mathsf d y}{\int_0^1\int_{0.9}^1 (x+y)~\mathsf d x~\mathsf d y}}$$

Then evaluate those two integrals and simplify the resulting ratio.

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  • $\begingroup$ I see! I was thinking that but wasn't really sure. But now I have another question: Why is $(P>0.9)$ also a double integral? Shouldn't it just be $\int_{0.9}^1 (x+y)~\mathsf d x$ since it's the marginal distribution of X? $\endgroup$ – dmalka May 19 '18 at 6:55
  • $\begingroup$ $\mathsf P(X>0.9) = \mathsf P(0.9<X\leqslant 1 ~\cap~ 0\leqslant Y\leqslant 1)$ ... and if you want to use the marinal distribution, remember that: $$\int_0^1\int_{0.9}^1 (x+y)\mathsf d x~\mathsf d y = \int_{0.9}^1 (x+\tfrac 12)\mathsf d x$$ $\endgroup$ – Graham Kemp May 19 '18 at 7:56

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