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Let $X$ and $Y$ be continuous random variables with joint density function

$$f(x,y) = \begin{cases}24xy& \text{for } x>0,\; y>0,\; 0<x+y<1\\ 0 &\text{otherwise} \end{cases}$$

What is the conditional probability $$P\left(X < \frac 12 \;\left|\; Y = \frac 14\right)\right.$$

Any help would be much appreciated. Thank you.

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Let $\left[x<\frac12\right]$ denote the function that takes value $1$ if $x<\frac12$ and takes value $0$ otherwise.

Then: $$\Pr\left(X<\frac12\mid Y=\frac14\right)=\frac{\int\left[x<\frac12\right]f(x,\frac14)dx}{\int f(x,\frac14)dx}\tag1$$

The function $g$ prescribed by:$$u\mapsto\frac{f(u,\frac14)}{\int f(x,\frac14)dx}\tag2$$ can be interpreted as the PDF of $X$ under condition $Y=\frac14$.

Observe that $(1)$ can also be written as: $$\int\left[x<\frac12\right]g(x)dx\tag3$$

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  • $\begingroup$ Thank you. But how do you evaluate P(Y=1/4)? $\endgroup$ – geniwebb Apr 19 '17 at 11:55
  • $\begingroup$ It is not necessary to evaluate $P(Y=\frac14)$. Working out the integral at the RHS of (1) is enough to find the mentioned conditional probability. If you are still curious: $P(Y=\frac14)=0$ because $Y$ is an (absolutely) continuous random variable. $\endgroup$ – drhab Apr 19 '17 at 12:07
  • $\begingroup$ So I need to integrate this one, ∫ 24xy g(x) dx ? $\endgroup$ – geniwebb Apr 19 '17 at 12:16
  • $\begingroup$ What are you trying to find by that integration? In order to find $\Pr(X<\frac12\mid Y=\frac14)$ you must find the RHS of (1) or (equivalently) (3). $\endgroup$ – drhab Apr 19 '17 at 12:21
  • $\begingroup$ P(X<1/2|Y=1/4) = P(X<1/2 union Y=1/4) / P(Y=1/4). This is my initial step in my computation. $\endgroup$ – geniwebb Apr 19 '17 at 12:22

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