Let the joint pdf of $X$ and $Y$ be $f(x,y) = 12e^{-4x-3y}, x>0, y>0$.

What is the marginal cdf of $X$? of $Y$?

Am I just supposed to integrate f(x,y) with respect to $x$ or $y$ to get the marginal cdfs?

Edit: Update with work $\begin{align*} f_1(x) = 12 \int_{0}^{\infty}e^{-4x-3y}dy = 12 \lim_{t\to\infty}\int_{0}^{\infty}e^{-4x-3y}dy = 12\lim_{t\to\infty}\bigg[ \frac{-1}{3}e^{-4x-3y}\bigg]_{0}^{t} = -4e^{-4x} \end{align*}

$\begin{align*} f_2(y) = 12 \int_{0}^{\infty}e^{-4x-3y}dx = 12 \lim_{t\to\infty}\int_{0}^{\infty}e^{-4x-3y}dx = 12\lim_{t\to\infty}\bigg[ \frac{-1}{4}e^{-4x-3y}\bigg]_{0}^{t} = -3e^{-4y} \end{align*}$

  • $\begingroup$ You don't actually have to integrate or take limits: simply observe that because $f$ factors into a constant, $e^{-4x},$ and $e^{-3y},$ the marginal PDF of $X$ must be proportional to $e^{-4x}$ and the marginal PDF of $Y$ must be proportional to $e^{-3y}.$ $\endgroup$
    – whuber
    Oct 31, 2019 at 20:30

1 Answer 1


You need to integrate against $y$ to find the $x$-marginal, and integrate against $x$ to find the $y$-marginal. We have

$$\begin{align*} f_X(x) &=\int_0^{\infty} 12e^{-4x-3y}dy\\\\ &12e^{-4x}\int_0^{\infty}e^{-3y}dy\\\\ &=12e^{-4x}\Big(-\frac{1}{3}e^{-3y}\bigg\rvert_0^{\infty}\Big)\\\\ &=12e^{-4x}\left(0-\left(-\frac{1}{3}\right)\right)\\\\ &=4e^{-4x} \end{align*}$$

so $X\sim\text{exp}(4)$. Similarly, you should find $Y\sim\text{exp}(3)$ and you can get the desired cdf's from there.

  • $\begingroup$ I got $f_X(x) = -4e^{-4x} and $f_Y(y) = -3e^{-4y}$, but this doesn't quite fit into the exponential distribution pdf $\frac{1}{\lambda} e^{-x/\lambda}$. I updated my work in the main post $\endgroup$
    – Evan Kim
    Oct 31, 2019 at 13:00
  • $\begingroup$ I have edited my answer, you made a slight mistake somewhere with your integration. $\endgroup$
    – Remy
    Oct 31, 2019 at 19:45
  • $\begingroup$ @EvanKim If you don’t get the heebie-jeebies upon getting PDFs that are negative-valued and think everything is A-OK, there is little hope that you can make progress on this question. $\endgroup$ Oct 31, 2019 at 22:50

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