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Suppose we have $0\le\,X\le\,\infty$, and we have $Y = a+b\, X\,\quad$ where $a\in\mathbb{R}$, $b\in\mathbb{R}$, $X \sim \chi^2_\nu(\beta)\quad$ with a PDF $f_X(x)$. Now, I found that the PDF of Y is given by

$f_Y(y) = \frac{1}{b}f_X\left(\frac{y-a}{b}\right)$

I have two questions please....

  • is the formula right for $f_Y(y)$.
  • if I want to obtain the expectation of another function of $Y$, given by $g(y)$, is it right to use

$\int^\infty_0 g(y)f_Y(y)\mathrm{d}y$

or $\int^\infty_a g(y)f_Y(y)\mathrm{d}y$.

Thanks.

1 Answers 1

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The pdf you give is correct for values of $z$ that are more than $a$, provided $b$ is positive.

You can prove that as follows: $ f_y(z) = \frac{d}{dz} F_y(z) = \frac{d}{dz} \Pr(y\le z) = \frac{d}{dz}\Pr(a+bx\le z) = \frac{d}{dz} \Pr\left( x \le \frac{z-a}{b} \right) $ where the very last step works only if $b>0$. Then: $ \frac{d}{dz} \Pr\left( x \le \frac{z-a}{b} \right) = \frac{d}{dz} F_x\left( \frac{z-a}{b} \right) = f_x\left( \frac{z-a}{b} \right) \cdot \frac 1b. $ In the last step, the chain rule is used.

For the expectation, you can use either $ \int_a^\infty g(z) f_y(z)\;dz $ or $ \int_0^\infty g(a+bx) f_x(w)\;dw. $ The second form given here is an instance of the "law of the unconscious statistician".

I prefer to use capital letters for random variables and often corresponding lower-case letters for the arguments to the cdf or pdf, thus: $ X \sim \chi^2_\nu $ $ \Pr(a < X < b) = \int_a^b f_X(x)\;dx. $

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    Thanks @Dilip Sarwate, so I can safely use the formula $\int^\infty_0g(y)f_Y(y)\,\mathrm{d}y$...2011-11-24