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So I took a test. And it was easy, but for some reason I got 0pts out of 5pts in a once single exercise.

OK, here it is: "Given the joint density function $$f(x,y)=12/7$$

when $$0\leq x \leq 1 \text{ and } 0\leq y \leq 1$$ And 0 elsewhere. Calculate it's marginal density of $X$."

The answers: (a): $$f_X(x)=(12x^2+6x)/7$$

(b): $$f_X(x)=(2x^2+3x)/7$$

(c): $$f_X(x)=(4+6x)/7$$

(d): $$\text{None of the above.}$$

My short solution sketch: Apply definition 5.4 (attached as image) and apply the bounds, hence $$f_X(x)=\int_0^1 12/7 dx=[12x/7]_0^1=12/7$$ thus, it's option D.

Definition 5.4 can be found here: http://puu.sh/BQaS3/5af75ea1be.png Am I right or? Because I did a similar exercise in the exercise classes where we have another number we should integrate (without variables) like here. I obtained a correct solution, but now I got 0 out of 5 points.

NabbKitha
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2 Answers2

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$f(x,y)$ doesn't integrate to $1$, it is not a valid joint density function.

Siong Thye Goh
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  • Yea, but I didn't make any typos, this is the question I got. But my problem is still that I got 0 out of 5pts. Anyway, are my method correct? I.e. my solution sketch? – NabbKitha Oct 24 '18 at 12:28
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    To get marginal distribution of $X$, you should integrate with respect to $y$. But it is really not a valid distribution, hence I would guess none of the above since the question doesn't make sense? – Siong Thye Goh Oct 24 '18 at 12:30
  • I would tell my instructor about it next time. Here is the question (I can't see the answers since the test is graded)

    http://puu.sh/BQbdO/36a5b21738.png

     – NabbKitha Oct 24 '18 at 12:31
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    most likely a typo somewhere in the question. – Siong Thye Goh Oct 24 '18 at 12:33
  • Thank you, I've sent an mail to my instructor and showed my solution. I've corrected the variable x to y, thanks for mention me about it. – NabbKitha Oct 24 '18 at 12:38
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I think the pdf is not right. I think the (valid) pdf is

$$f(x,y)=12/7\cdot (x^2+xy)$$

when $0\leq x \leq 1 \text{ and } 0\leq y \leq 1$

To calculate the marginal density of $X$ you have to integrate w.r.t $y$- $\color{blue}{\texttt{not x}}$

$$f_X(x)=\int_0^1 12/7\cdot (x^2+xy) \, dy $$

$$ \frac{12}{7} \cdot \left[ x^2y+\frac{1}{2}xy^2 \right]_0^1=\frac{12}7\cdot (x^2+\frac12 x)-0$$

$$=\frac{1}{7} \cdot (12x^2+6x)$$

when $0\leq x \leq 1 $

Therefore the answer is a)

callculus42
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