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I am working on the 6.041 MIT Opencourseware course. Though, currently I am struggling with assignment 1b of the attached assignment (see pic).

As the joint pdf is linear (red circle), I do not see why the marginal PDF of Y can be quadratic (orange arrow). My feeling is that the provided solution mistakenly calculates the CDF.

What I did when trying to solve this; was to create a definite integral with bounds 1 and 2. In other words, replace the y (green arrow) with a 2. Though, as a result I would then have a new problem, as with this approach I would only have a constant (no variable) in my marginal PDF.

Do I have a mistake in my understanding? Thanks in advance!

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  • $\begingroup$ It would be beneficial if you could typeset your question plus your attempts in MathJax. General hint: The marginal pdf of a joint pdf is obtained by integrating out the other variable. Here $x$. $\endgroup$
    – Kurt G.
    Feb 4, 2023 at 16:23
  • $\begingroup$ Please format your question using Mathjax $\endgroup$ Feb 4, 2023 at 19:49
  • $\begingroup$ I think you misremembered what "marginal PDF" means. $\endgroup$ Feb 6, 2023 at 15:37

1 Answer 1

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First, draw the three inequalities $$ 1 \leq x \leq y \leq 2$$ in the xy-plane. Then find their intersection. enter image description here

To find the marginal PDF $f_Y(y)$, you have to integrate $f_{X,Y}(x,y)$ with respect to $x$ over the triangle depicted in the figure. Look at the arrow. It passes through $x=1$ first and then it passes through $x=y$. So $$ f_Y(y)=\int_{x=1}^{x=y} \frac{3}{2}x\, dx$$ for $1\leq y \leq 2$.

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  • $\begingroup$ Thank you so much! It clicked! $\endgroup$
    – Ewout
    Feb 5, 2023 at 10:19

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