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Let $X_1$ denote the weight (in tons) of a bulk item stocked by a supplier at the beginning of a week and suppose that $X_1$ has a uniform distribution over the interval $0 \le X_1 \le 1$. Let $X_2$ denote the amount (by weight) of this item sold by the supplier during the week. Given the value $X_1=x_1$, $X_2$ has a uniform distribution over the interval $[0,x_1]$, where $x_1$ is a specific value of $X_1$.

a)Find the joint density function of $X_1$ and $X_2$.

b.If the supplier stocks a half-ton of the item, what is the probability that she sells more than a quarter ton?

c. If it is known that the supplier sold a quarter ton of the item, what is the probability that she had stocked more than a half-ton?


Since the two variables are not indepent, I find it hard to say anyting about their joint density function. What I first of all did was find the PDF for $X_1$ which is obviously just 1. For $X_2$ the PDF should be $1/x_1$, I would say. However, how do these help me to find the joint density function?

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Yes you're right for c), you need to calculate $$ P(X_1 > \frac 1 2 \vert X_2 = \frac 1 4). $$ Ok, you know that $X_2 \vert X_1$ is uniformly distributed on $(0,X_1)$ so $$ P(X_2 > \frac 1 4 \vert X_1 = \frac 1 2) = P( \mathcal U(0,\frac 1 2) > \frac 1 4) = \int_{1/4}^{1/2} 2 dx = \frac 1 2. $$ For c), calculate the density of $X_2 \vert X_1$ using Bayes formula and do the same.

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  • $\begingroup$ For c, I think Bayes formula gives you f(X1|X2)=f(X1)f(X2|X1)/f(X2) = 1 * (1/x1) / f(X2). However, how do I find f(X2)? Or am I doing something wrong? Thanks a lot again :). $\endgroup$ – dreamer Mar 24 '13 at 11:43
  • $\begingroup$ Could you please help me out on c? I still cant seem to find the correct answer. Integrating 1/x1 over x1 to find f(x2) doesnt give me a finite answer. How should I do this instead? @roger $\endgroup$ – dreamer Mar 25 '13 at 16:02
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You know the distribution of $X_2 \vert X_1$ and that of $X_1$ so you can deduce their joint distribution.

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  • $\begingroup$ Are you referring to the formula for conditional probabilities? So you mean that P(X1 intersection X2) = P(X2|X1)*P(X1)? Does this work for density functions too? Or is this not what you meant? $\endgroup$ – dreamer Mar 22 '13 at 19:46
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    $\begingroup$ Yes this is what I mean. Bayes formula does work for density too $\endgroup$ – roger Mar 22 '13 at 19:48
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    $\begingroup$ Yes, on the condition that $0 < x_2 < x_1 < 1$. It's zero otherwise. $\endgroup$ – TakeS Mar 22 '13 at 21:22
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    $\begingroup$ actually it's $1/x_1$ if $0\leq x_2 \leq x_1 \leq 1$ and $0$ elsewhere. $\endgroup$ – roger Mar 22 '13 at 21:22
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    $\begingroup$ for b), you need to calculate $$ P(X_2 > \frac 1 4 \vert X_1= \frac 1 2) $$ for c) $$ P(X_1 > \frac 1 2 \vert X_2 > \frac 1 4) $$ for b), simply use the density of $X_2 \vert X_1$ and for c) use Bayes's formula. $\endgroup$ – roger Mar 23 '13 at 13:24

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