$f(x,y) = 1.5x$ for $0 \leq x\leq 1$ and $0 \leq y \leq 2x$.

I want to find the pdf.

I tried integrating $1.5x$ from $0$ to $1$ with respect to $x$ and got $f(y)=\frac{3}{4}$ for $0\leq y\leq 2$. However, this is clearly not a valid pdf, as $\int_0^2\frac{3}{4}dy = 1.5 \neq 1$.

Can anyone see my error?

  • 1
    $\begingroup$ You need to take into account the fact that $x$ determines the support of $y$. Seeing it in reverse, how this affects the interval over which you must integrate-out $x$? $\endgroup$ – Alecos Papadopoulos Aug 30 '18 at 16:54
  • $\begingroup$ To supplement what @Alecos said: draw a picture. $\endgroup$ – whuber Aug 30 '18 at 17:00

The error might be clearer if we actually look at a specific value of $y$, say $y=1$. According to your method $$f(1) = \int_0^1 f(x, 1) \mathrm{d}x$$ This integral includes, for example, $f(1/4, 1)$. But then you replace $f(1/4, 1)$ with $3/8$ ($1.5x$ evaluated at $x=1/4$) when in fact if $y=1$ then $x$ must be at least $1/2$, so $f(1/4, 1)$ is zero.

Looking at the domain of $y$, we see that $0\leq y \leq 2x$ which implies that $x\geq y/2$. So for $x\leq y/2$ should be ignored when we calculate $f(y)$. Changing the limits of the integral to reflect this gives us

$$f(y) = \int_{y/2}^1 f(x,y)\mathrm{d}x = \int_{y/2}^1 1.5x\mathrm{d}x$$ Computing this integral gives a function of $y$ for the marginal $f(y)$, if you integrate this again over $y$ it will be $1$.

Essentially you've used the expression $f(x, y) = 1.5x$ without checking that the point $(x, y)$ is a valid one.


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