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Let the continuous random variables $X$ and $Y$ have the joint probability density function given by $f(x)$ = $kx$ for $0<x<2$, $0<y<1$, $x<2y$. Find $k$.

I found the joint probability and equated it to one to $k=3/2$. Could that be the right answer? Where $x$ is integrated from $2y$ to $2$ and $y$ from $0$ to $1$.

What is the conditional probability density function $f(x|y)$?

I defined $f(x|y)$ = $f(x,y)$/$f(y)$. But don't know how to go about it?

J.R.
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1 Answers1

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$\checkmark$ Looks all right so far. (Although $x$ needs to be integrated from $0$ to $2y$.)

You'll next need to use: $$\begin{align} f(y) & = \int_0^{2y} f(x,y)\operatorname d x \cdot\operatorname{\bf 1}_{(0,1)}(y) \\[1ex] & = \frac 3 2 \int_0^{2y} x \operatorname d x\cdot\operatorname{\bf 1}_{(0,1)}(y) \\[3ex] & = 3 y^2\cdot\operatorname{\bf 1}_{(0,1)}(y) \\[2ex]\hline f(x\mid y) & = \frac{f(x,y)}{f(y)} \\[1ex] & = \frac{x}{2 y^2}\cdot\operatorname{\bf 1}_{(0,1)}(y)\cdot\operatorname{\bf 1}_{(0,2y)}(x) \end{align}$$

Graham Kemp
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  • @Thanks Graham. Does it suffice to say that $f(x)$ = 3/2x? – J.R. Nov 08 '14 at 17:04
  • I tried finding $P(x<1.5|y=0.5)$ but got 2.25. That doesnt appear right since the probability cannot be greater than one. What am I doing wrong? – J.R. Nov 08 '14 at 20:43
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    @J.R. Did you consider the support for the conditional density function? $\ \mathsf P(x<1.5\mid y=0.5) = \int_0^1 \frac{ x }{2(0.5)^2} \operatorname d x = 1$; because $y=0.5 \land x\in(0,2y)\Rightarrow x\in(0,1)$ – Graham Kemp Nov 08 '14 at 22:29
  • I forgot to change the limits of integration. But your input helped. – J.R. Nov 08 '14 at 23:37
  • I want help with this question.

    Find the probability density function $U=X-Y$. Use figures to help you find $f(u)$.

    I approached this problem using transformation and had $0<u<2$. I need to find the distribution function for $0<u<1$ and $1<u<2$. But the limits of integral is my headache. Is there a better way in deriving the limits.

     – J.R. Nov 08 '14 at 23:37