2

Consider $f_X(x) = 1/2+x$ where $0 <x\leq 1$. We want to find the pdf of $Y=X^{-1}$ by utilizing the following transformation.

$$f_Y(y) = \int_0^1 f_X(x) \delta (y - g(x)) dx$$

Where $u =g(X) = Y$.

From my understanding we can represent $x$ as $x=g^{-1}(g(x))$ and then we can write $dx$ as $dx = \frac{d(g(x))}{dx}dx \implies dx=-x^2du$. Since $x \in (0,1]$, we can conclude that $y \in [1,\infty)$. So we can rewrite our expression as:

$$f_Y(y) = \int_1^\infty f_X(g^{-1}(g(x))) \delta (y - g(x)) (-\frac{1}{y^2})du$$ However, after this point I don't really know how to continue. Yes I can factor out a negative sign, but i just don't see what my next step is. Any ideas?

WaterDrop
  • 600

2 Answers2

1

Are you aware of the filtering/sifting property of Dirac Delta function? The filtering property says that $\forall\epsilon,\gamma>0$ the value of$$\int_{a-\epsilon}^{a+\gamma} h(x)\delta(a-x)dx=\lim_{x\to a}h(a)$$ Take $g(x)=1/x$ as $m$, then you have$$f_Y(y)=\int_1^\infty\frac1{m^2}f_X(1/m)\delta(y-m)~dm$$

For $y<1,$ $\delta(y-m)=0$ so $f_Y(y)=0,y<1$. For $y>1$, we have $1/y\in(0,1)$ and so$$f_Y(y)=\lim_{m\to y}\frac{f_X(1/m)}{m^2}=\lim_{1/m=k\to1/y}k^2f_X(k)=\frac{f_X(1/y)}{y^2}$$since $k^2f_X(k)$ is continuous in $(0,1)$. Finally you get$$f_Y(y)=\begin{cases}0,&y\le1\\\frac1{y^2}\left(\frac12+\frac1y\right),&y>1\end{cases}$$

Shubham Johri
  • 17,659
0

\begin{align} &\int_{0}^{1}\operatorname{f}_{X}\left(x\right) \,\delta\left(y - {1 \over x}\right)\mathrm{d}x = \int_{0}^{1}\operatorname{f}_{X}\left(x\right) \,{\delta\left(x - 1/y\right) \over \left\vert\,{ 1/x^{2}\,}\right\vert}\mathrm{d}x \\[5mm] = &\ \left.\vphantom{\Large A} \left[0 < {1 \over y} < 1\right] x^{2}\operatorname{f}_{X}\left(x\right) \,\right\vert_{\, x\ =\ 1/y}\,\,\,\,\, = \,\,\,\,\, \bbox[10px,border:1px groove navy]{\left[y > 1\right]\,{\operatorname{f}_{X}\left(\,{1/y}\,\right) \over y^{2}}} \\ & \end{align}

Felix Marin
  • 89,464