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Find the marginal distribution for $y_2$ given the following PDF

$$f(y_1,y_2)= \begin{cases} 3y_1, & \text{if } 0\leq y_2\leq y_1 \\ 0, & \text{elsewhere} \end{cases} $$

But when I try to find it, I end up with the following integral:

$$f_{Y_2}(y_2)=\int_{y_2}^\infty 3y_1 \, dy_1$$

and that integral does not converge. Is there any trick to solve this? or is this situation an special case?

Thanks

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  • $\begingroup$ You should have $f_{Y_2}(y_2)$ where you have $f_{y_2}(y_2). \qquad$ $\endgroup$ Commented Aug 26, 2020 at 23:17
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    $\begingroup$ The function that you called the PDF is not in fact a probability density function, for just the reason you stated. $\endgroup$ Commented Aug 26, 2020 at 23:18
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    $\begingroup$ This is not a PDF as it stands. I suspect you mean $0 \le y_2 \le y_1 \le 1$. $\endgroup$ Commented Aug 26, 2020 at 23:19

1 Answer 1

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Assuming you actually meant $0<y_2<y_1<1$, first check that you have pdf by integrating: $$ \int_0^1 \int_0^{y_1} 3y_1 \, dy_2 \, dy_1 $$ Once you've done that, marginalize out $Y_2$: $$ f_{Y_2}(y_2) = \int_{y_2}^1 3y_1 \, dy_1 $$ Again, check this by integrating in the $[0,1]$ interval to get $1.$

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  • $\begingroup$ No, in my homework is $0\leq y_2\leq y_1$. I think the way my teacher write the homework is wrong $\endgroup$ Commented Aug 26, 2020 at 23:49
  • $\begingroup$ Try integrating the joint pdf, and see if you can get $1$. If not, there's an error somewhere $\endgroup$
    – Alex
    Commented Aug 26, 2020 at 23:51

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