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I want to find the probability of $P(X<0.5 \;|\; Y>0.25)$ of a joint pdf

$$f_{(X,Y)}(x,y)= \frac 1 x $$

with $0<y<x<1.$

This should be as follows:

$$P(X<0.5\;|\; Y>0.25)=\frac{P(X<0.5 \;,\; Y>0.25)}{P(Y>0.25)}$$

The denominator can be calculated as

$$1-P(Y\leq0.25)=1-\int_{y=0}^{0.25}f_Y(y) dy =1-\int_{0}^{0.25} -\ln y \;dy=1-0.596$$

since $f_Y(y)=\int_{y}^1 1/x \; dx=-\ln(y).$

As for the numerator:

$$P(X<0.5 \;,\; Y>0.25)=\int_{x=0}^{0.5}\int_{y=0.25}^{x}f_{X,Y}(x,y) \; dy dx=\int_{x=0}^{0.5}\Big[y/x\Big]^{x}_{0.25}\;dx\\=\int_{x=0}^{0.5}1-0.25/x\;dx=\Big[x- 0.25 \ln x\Big]^{0.5}_0=0.5-0.25 \ln(0.5)+0.25 \ln(0)$$

but $\ln(0)$ is $-\infty$ and the integral does not converge. What have I done wrong?

I see this works changing the limits of integration to

$$P(X<0.5 \;,\; Y>0.25)=\int_{x=0.25}^{0.5}\int_{y=0.25}^{x}f_{X,Y}(x,y) \; dy dx$$

which makes sense looking at the area of integration, but having factored the relationship between $x$ and $y$ in the limits of the inner integral, I don't understand why we can't run the limits from $0$ to $0.5$ on the outside integral.

JAP
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1 Answers1

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Your issue is the logic in the numerator:

$P(X<0.5,Y>0.25)$

if $0<y<x<1 $ , then $P(X<0.5,Y>0.25) = P(Y>0.25, X>0.25, X<0.5)$

which is:

$P(Y>0.25, 0.25<X<0.5)$

That explains why you cannot run the integral of x between 0 to 0.25 and therefore:

$$P(0.25<X<0.5 \;,\; Y>0.25)=\int_{x=0.25}^{0.5}\int_{y=0.25}^{x}f_{X,Y}(x,y) \; dy dx$$

Remark: I just rearranged the order of the random variables above, to attempt to make it easier to understand.

Rahul P
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