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The joint density of Y1 , the proportion of the capacity of the tank that is stocked at the beginning of the week, and Y2 , the proportion of the capacity sold during the week, is given by

f(y1,y2)={
          3*y1, if 0 ≤ y2 ≤ y1 ≤ 1,
          0, elsewhere
         }

Question: Find P(Y2 ≤ Y1/2), the probability that the amount sold is less than half the amount purchased.

I am not sure how to set up the integral for the limits Y2 ≤ Y1/2 in P(Y2 ≤ Y1/2). I know that 3*y1 is bounded between 0 ≤ y2 ≤ y1 ≤ 1 so that forms a triangle beneath the line y = x with base and height equal to 1.

But for Y2 ≤ Y1/2 do I draw a separate graph, or do I bound the Y2 ≤ Y1/2 into that same graph of 0 ≤ y2 ≤ y1 ≤ 1 which is given in the original question.

I just need a starting point. I am not sure where to begin in this question.

CapturedTree
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1 Answers1

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Start with one variable, $y_1$, ignoring the restrictions from $y_2$ examine the restrictions on the variable $y_1$. We have $0\leq y_1 \leq 1$.

Now that you have details of $y_1$ you can examine $y_2$ in terms of $y_1$. Given the value of $y_1$ what are the restrictions on $y_2$? We have $0 \leq y_2 \leq y_1$ from the first part of the question and $y_2 \leq \frac{y_1}{2}$ from the second part of the question. Combining these two requirements on $y_2$ gives $0 \leq y_2 \leq \frac{y_1}{2}$

Therefore the integral is

$\int_{y_1=0}^{1} \left(\int_{y_2=0}^{\frac{y_1}{2}} f(y_1,y_2)dy_2 \right) dy_1$

Notice that the limits of $y_2$ contain information about $y_1$ so the integration with respect to $y_2$ has to be done first.

Hugh
  • 2,341
  • Ohh ok that makes sense. Now using the same logic if I have the limits y1 > 0 and y2 > 0 and was asked to find P(y1 + y2 < 3). Would be limits be 0 to infinity for dy1 and for dy2 since y2 > 0 and since y2 < 3 - y1 the limits would be 0 to 3 - y1 for dy2? – CapturedTree Nov 23 '16 at 22:49
  • @1290 Not quite, the limits $0< y_2< 3-y_1$ is correct but $y_1$ cannot be greater than 3 – Hugh Nov 23 '16 at 22:55
  • Wait shouldn't it be 0 to 3 - y2 for dy1? Since y1 > 0 and from y1 + y2 < 3 we get y1 < 3 - y2 so that implies 0 to 3 - y2 for dy1? (0 < y1 < 3 - y2) – CapturedTree Nov 23 '16 at 23:00
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    @1290 You can do it that way too. There's a symmetry between $y_1$ and $y_2$ in the question so the limits work both ways. – Hugh Nov 23 '16 at 23:25
  • Oh ok. Thanks for all the help I am starting to understand this a lot better. – CapturedTree Nov 23 '16 at 23:29