Sum of a random variable and its own square

Question

So I have the random variable $X$ in the following function:

$$ g(X) = \frac{X}{T}\left[ 1 + (X-1)\text{sinc}^{2}(fT) \right] $$

Expanding, gives us:

$$ g(X) = \frac{1 - \text{sinc}^{2}(fT)}{T}X + \frac{\text{sinc}^{2}(fT)}{T}X^{2} $$

How can I find the distribution of $Y = g(X)$, given that it is clearly not monotonic?

Note: My random variable $X$ can be Gamma, Erlang, or Chi distributed, and $f$ and $T$ are constants.

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Edit: I remember something about using delta functions, but I cannot find the reference now...

Answer 1

Set $a=\frac{\text{sinc}^2(fT)}{T}$ and $b=\frac{1-\text{sinc}^2(fT)}{T}$ . Suppose $X$ is supported on $[0,\infty)$ (which is the case for Chi-squared, Erlang, and Gamma distributions). Assume $a,b>0$ and fix $y\geq 0$. Notice $$aX^2+bX\leq y \iff X\in\Bigg[\frac{-b-\sqrt{b^2+4ay}}{2a},\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg]$$ Since $P\Bigg(\frac{-b-\sqrt{b^2+4ay}}{2a}\leq X <0\Bigg)=0$ we have that $$F_{Y}(y)=P(Y\leq y)=P\Bigg(X\leq\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)=F_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)$$ Hence $$f_Y(y)=f_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)\frac{d}{dy}\Bigg[\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg]=f_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)\frac{1}{\sqrt{b^2+4ay}}$$

On the other hand, if $a>0$ and $b<0$, we see $Y$ is supported on the interval $\big[-\frac{b^2}{4a},\infty\big)$. If $-\frac{b^2}{4a} \leq y \leq 0$ then

$$F_{Y}(y)=P(Y\leq y)=P\Bigg(\frac{-b-\sqrt{b^2+4ay}}{2a} \leq X \leq \frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)=F_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)-F_X\Bigg(\frac{-b-\sqrt{b^2+4ay}}{2a}\Bigg)$$ Now if $y>0$ then

$$P(Y \leq y)=P\Bigg(X \leq \frac{-b+\sqrt{b^2+4ay}}{2a} \Bigg)=F_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)$$

So we get $$f_Y(y)=\frac{1}{\sqrt{b^2+4ay}}\Bigg[f_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)+f_X\Bigg(\frac{-b-\sqrt{b^2+4ay}}{2a}\Bigg)\Bigg] $$ whenever $-\frac{b^2}{4a} < y \leq 0$ and $$f_Y(y)=\frac{1}{\sqrt{b^2+4ay}}f_X\Bigg(\frac{-b+\sqrt{b^2+4ay}}{2a}\Bigg)$$ whenever $y > 0$. Now let's consider the case when $a=0,b>0$. This is only possible if $Y=X/T$ which is supported on $[0,\infty)$ so for $y\geq 0$ we have $$F_{Y}(y)=P(Y \leq y)=P(X\leq Ty)=F_X(Ty)$$ This means $$f_Y(y)=f_X(Ty)\frac{d}{dy}\Big[Ty\Big]=Tf_X(Ty)$$ Lastly, if $a>0 \text{ and } b=0$, then $Y=X^2/T$ which is supported on $[0,\infty)$. So for $y \geq 0$ we have $$F_{Y}(y)=P(Y \leq y)=P\big(X^2 \leq Ty\big)=P\big(X \leq \sqrt{Ty}\big)=F_X\Big(\sqrt{Ty}\Big)$$ Finally, $$f_{Y}(y)=f_X\big(\sqrt{Ty}\big)\frac{d}{dy}\Big[\sqrt{Ty}\Big]=\frac{1}{2}\sqrt{\frac{T}{y}}f_X\big(\sqrt{Ty}\big)$$ for $y>0$. All of these pdf's for $Y$ vanish elsewhere.

Answer 2

Since the coefficients on the $X$ and $X^2$ are constants, more generally, we can find the distribution of $$Y \sim aX^2+bX$$ The probability that $Y = y$ is given by the probability that $aX^2+bX = y$. This is equal to the probability that $aX^2+bX - y = 0 \to X = \frac{-b \pm \sqrt{b^2+4ay}}{2a}$. This means that the probability $Y = y$ is given by $$P\left(X=\frac{-b - \sqrt{b^2+4ay}}{2a}\right)+P\left(X=\frac{-b + \sqrt{b^2+4ay}}{2a}\right)$$

Edit: The above probability would technically be $0$ for a continuous probability distribution for a given $y$. This is actually the PDF of $y$.

I wasn't being very mathematically rigorous previously, but the CDF would be $$P(Y < y) = \int_{-\infty}^{y} f_Y(t) dt$$ where $f_Y(t)$ is the PDF of $Y$. The PDF is given by $P(Y=y)=f_Y(y)$.