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I have the joint density function $f(y_1,y_2) = 3y_1, 0 \leq y_2 \leq y_1 \leq 1.$ And $0$ elsewhere.

The conditional density for $f(y_1|y_2)$ is defined over $y_2 \leq y_1 \leq 1$, provided that $y_2 \leq 0 $ .

The question is about Y1, the proportion of items stocked and Y2, proportions of items sold.

I have to find the probability that more than 1/2 is sold given that 3/4 is stocked.

My atempt is to define $f(y_1|y_2)$:

$f(y_1|y_2)= \frac{f(y_1,y_2}{f_2(y_2)} = \frac{1/2}{3/4 y_2} = \frac{2}{3y_2}$

Then calculate $P(y_1 > 1/2 \; | \; 3/4) $ With the integral $\int_{\frac{1}{2}}^{1} \frac{2}{3y_2} dy = \frac{1}{6}$.

But the result should be 1/3 according to my textbook, so I'm doing something wrong somewhere. Hope someone can help me with figuring out where I'm going wrong.

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  • $\begingroup$ The approach is the right one but you confused $y_1$ and $y_2$ (and you didn't compute correctly the marginal density). $\endgroup$
    – Papi
    Commented Nov 8, 2021 at 11:07

1 Answer 1

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Hope someone can help me with figuring out where I'm going wrong.

As you wrote, your question is about $P(Y_2|Y_1=y_1)$ thus the conditional density you found is not useful...and even wrong

Just to simplify the notation, let's set

$X=$ stocked goods

$Y=$ sold goods

First calculate

$$f_{Y|X}=\frac{f_{XY}}{f_X}=\frac{3x}{3x\int_0^x dy}=\frac{1}{x}$$

thus $(Y|X=x)\sim U(0;x)$

given that $x=0.75$ you have that

$$f_{Y|X=0.75}(y)=\frac{4}{3}\cdot\mathbb{1}_{(0;0.75]}(y)$$

And concluding,

$$\mathbb{P}[Y>0.5|x=0.75]=\left[\frac{3}{4}-\frac{1}{2}\right]\times\frac{4}{3}=\frac{1}{3}$$

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