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I'm having trouble with the following random variable transformation:

$Y = X^2 + X$

I am looking for the pdf of Y. I tried the following method:

$p_Y(y) = \int_{X} p_{Y|X=x}(y)\cdot p_{X}(x)dx$ and we know that $(Y|X=x) \sim (x^2+x) \Rightarrow p_{Y|X=x} = \delta_{x^2+x}(y)$ thus: $p_Y(y) = \int_{X} \delta_{x^2+x}(y)\cdot p_{X}(x)dx$

But I don't see a way to reduce this further.

Then I tried a different approach:

$p_Y(y) = DF(Y < y) = DF(X^2 + X < y) = ...$

But then I don't see a way to find the inverse of $X^2 + X$.

Can anyone help me further on this?

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Let $Y = X^2 + X = \left( X+\frac{1}{2} \right)^2 - \frac{1}{4}$. Then $$ F_Y(y) = \mathbb{P}(Y \le y) = \mathbb{P}\left( \left( X+\frac{1}{2} \right)^2 \le y + \frac{1}{4} \right) $$ Assume, additionally, that $y+\frac{1}{4} > 0$. Then $$ \begin{eqnarray} F_Y(y) &=& \mathbb{P}\left( -\sqrt{y + \frac{1}{4}} \le X+\frac{1}{2} \le \sqrt{y + \frac{1}{4}} \right) \\ &=& F_X\left( -\frac{1}{2} + \sqrt{y + \frac{1}{4}} \right) - F_X\left( -\frac{1}{2} -\sqrt{y + \frac{1}{4}} \right) + \mathbb{P}\left( X = -\frac{1}{2} - \sqrt{y + \frac{1}{4}} \right) \end{eqnarray} $$

If $X$ is continuous rv, the last term is zero.

Differentiating with respect to $y$, we get $$ f_Y(y) = \frac{1}{\sqrt{4y+1}} \left( f_X\left(-\frac{1}{2} + \sqrt{y + \frac{1}{4}} \right) + f_X\left( -\frac{1}{2} -\sqrt{y + \frac{1}{4}} \right) \right) \cdot \mathbf{1}\left(y > -\frac{1}{4}\right) $$

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Thanks for the hint. Indeed, when Y = X^2 + X, this reduction works, but actually I was looking for a way to transform a general case X^n + X^n-1 + ... + 1. But I suppose that will not be possible, as that function will not always be invertable –  James Nov 23 '11 at 21:07
    
$Y = X^n + X^{n-1} + \ldots + 1 = \frac{X^{n+1}-1}{X-1}$. Notice that the polynomial in $X$ only has 1 real root for even $n$ and no real roots for odd $n$. I think it is possible to work the solution out, but it will involve symmetric function of complex roots. –  Sasha Nov 23 '11 at 21:14
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