Suppose I have $X\sim U(-1,1)$ and $Z\sim U(0,0.1)$ independently and want to work out the pdf of a new r.v. defined as such: $Y=X^2+Z$. How could I calculate its distribution? I have tried to work it out but seem to have contradictory things appearing.

I have been able to show that the distribution of $Y|X=x \sim U(x^2,x^2+0.1)$ and hence we can deduce the joint pdf of $X$ and $Y$ is given by $$f_{XY}(x,y)=f_{Y|X}(y|x)f_X(x)=10\, \cdot \frac{1}{2}=5$$ and this is valid over the support $\{(x,y):-1<x<1,\, x^2<y<x^2+0.1\}$.

Hence to find $f_Y$ all that I surely need to do is integrate the joint pdf above over its support of $X$, ie. $\{x:-1<x<1\}$.

Doing this we get that $f_Y(y)=10$, which is valid for $\{y:0<y<1.1\}$, but how can this be? Where is the error in my reasoning?

  • 1
    $\begingroup$ $X^2$ has support on $[0,1]$ while $Z$ has support on $[0,0.1]$ so it might be sensible to analyse the distribution of $Y$ for the cases $y \lt 0$, $0 \lt y \lt 0.1$, $0.1 \lt y \lt 1$, $1 \lt y \lt 1.1$ and $1.1 \lt y$. The first and last are trivial, but the density is not constant in $[0,1.1]$ $\endgroup$ – Henry Apr 12 '18 at 15:27
  • $\begingroup$ There is no problem having the density greater than $1$ in places, so long as its overall integral across all real values is $1$ $\endgroup$ – Henry Apr 12 '18 at 15:29
  • $\begingroup$ $Z$ has support on $[0,0.1]$! Small correction @Henry $\endgroup$ – user258521 Apr 12 '18 at 15:30
  • $\begingroup$ user258521 - corrected $\endgroup$ – Henry Apr 12 '18 at 15:31
  • $\begingroup$ But then if the pdf is equal to $10$ across $(0,1.1)$ then isnt the overall integral over the support of $Y$ not equal to $1$? So if the density isn't constant over the interval then my pdf candidate is wrong? $\endgroup$ – user258521 Apr 12 '18 at 15:32

I prefer to handle cumulative distribution functions, rather like this:

For $0\le y \le 0.1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^y 10\, \mathbb P(X^2 \le y-z) \,dz = \int_{z=0}^y 10\, \sqrt{y-z} \,dz = \frac{20}{3}y^{3/2}$$

and taking the derivative will give you $f_Y(y)=10 \sqrt{y}$.

For $0.1\le y \le 1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^{0.1} 10\, \mathbb P(X^2 \le y-z) \,dz = \frac{20}{3}y^{3/2} - \frac{20}{3}(y-0.1)^{3/2}$$

and taking the derivative will give you $f_Y(y)=10 \sqrt{y}-10 \sqrt{y-0.1}$.

For $1\le y \le 1.1$ you can say $$\mathbb P(Y \le y) = \int_{z=0}^{y-1} 10 \,dz + \int_{z=y-1}^{0.1} 10\, \mathbb P(X^2 \le y-z) \,dz = 10y -\frac{10}{3} - \frac{20}{3}(y-0.1)^{3/2}$$

and taking the derivative will give you $f_Y(y)=10-10 \sqrt{y-0.1}$.

The cumulative distribution function turns out to look like

enter image description here

and the probability density function (zero outside $[0,1.1]$) looks like

![enter image description here


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