Let $X \sim U(0,1) $ and $Y \sim U(0,x) $. Calculate $$ \Pr(X >0.5 | Y= 0.25)$$
Is this a trick question ? Since $\Pr(Y = 0.25) = 0$, right ?
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3It is not supposed to be a trick question, though would need careful definition to be meaningful as you are correct that $\frac{\Pr(X >0.5, Y= 0.25)}{\Pr(Y= 0.25)}$ is not well-defined. Try $\lim\limits_{\varepsilon \to 0^+} \Pr(X >0.5 \mid 0.25 - \varepsilon < Y <0.25 + \varepsilon)$ – Henry Aug 19 '22 at 17:04
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1See https://stats.stackexchange.com/a/584907/919 for a discussion of this issue. – whuber Aug 19 '22 at 21:29
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$Y \sim U(0,x)$ is really saying that if $X$ has value $x \in (0,1)$, then the conditional pdf of $Y$ is uniform on $(0,x)$. That is, for $0 < y < x < 1$, $$f_{X,Y}(x,y) = f_{Y\mid X}(y\mid X=x)f_X(x) = \frac 1x \cdot 1 = \frac 1x, 0 < y < x < 1.$$ Note that from this we can find $f_Y(y)$ as $$f_Y(y) = \int_y^1 f_{X,Y}(x,y) \mathrm dx = \int_y^1 \frac 1x \mathrm dx = -\ln y, 0 < y < 1.$$ What you really should find is the conditional pdf of $X$ given that $Y=0,25$ from which calculation the desired quantity $P(X>0.5 \mid Y = 0.25)$ (which we recognize as the complementary conditional CDF of $X$ given that $Y = 0.25$) ir readily obtained. We have that $$f_{X\mid Y = 0.25}(x\mid Y=0.25) = \dfrac{f_{X,Y}(x,0.25)}{f_Y(0.25)}.$$ Can you finish from here?
Dilip Sarwate
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Is it $$\frac{\int_{0.5}^1 \frac{1}{x}dx}{- \ln{0.25}} = \frac{\ln{2}}{2\ln{2}} = \frac{1}{2} $$ – actsci stud tries2learn math Aug 20 '22 at 12:30