(New page: Let <br/> <math>X = X_1 - X_2 / </math><br/> where <math>X_1</math> and <math>X_2</math> are iid scalar random variables. Also, let <math>Y</math> be a chi-squared variable with 1 degree ...) |
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| − | + | Given <br/> | |
| − | <math>X = X_1 - X_2 | + | <math>X = X_1 - X_2 \ </math><br/> |
where <math>X_1</math> and <math>X_2</math> are iid scalar random variables. | where <math>X_1</math> and <math>X_2</math> are iid scalar random variables. | ||
| − | Also, | + | Also, Given that <math>Y</math> is a chi-squared variable with 1 degree of freedom. So,<br/> |
<math>f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}}</math> | <math>f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}}</math> | ||
| − | Additionally, | + | Additionally, it is given that <math>Y= X^2</math><br/> |
| + | |||
| + | We first make the following assumptions: | ||
| + | * <math>f_{X_1}(x)</math>, and hence <math>f_{X_2}(x)</math> are even functions. | ||
| + | * the Fourier transforms of <math>f_{X_1}(x)</math> and <math>f_{X_2}(x)</math> exist. | ||
| + | |||
| + | then we want to show that <br/> | ||
| + | |||
| + | <math>X_1, X_2 \sim N(\mu,\frac{1}{2}) \Leftrightarrow f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}}</math> | ||
| + | <br/> | ||
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| + | |||
| + | ---- | ||
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| + | To prove sufficiency, | ||
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| + | If <math>X_1</math> and <math>X_2</math> are two Gaussian distributed scalar random variables with mean <math>\mu</math> and variance <math>\sigma^2</math>, | ||
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<math>E[e^{-jty}] = E[e^{-jtx^2}] = \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx </math> | <math>E[e^{-jty}] = E[e^{-jtx^2}] = \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx </math> | ||
| + | |||
| + | We want to show that <br/> | ||
| + | |||
| + | <math>X_1, X_2 \sim N(\mu,\frac{1}{2}) \Leftrightarrow f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}}</math> | ||
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Let <math>u = x^2</math>, then<br/> | Let <math>u = x^2</math>, then<br/> | ||
Revision as of 10:42, 8 June 2013
Given
$ X = X_1 - X_2 \ $
where $ X_1 $ and $ X_2 $ are iid scalar random variables.
Also, Given that $ Y $ is a chi-squared variable with 1 degree of freedom. So,
$ f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}} $
Additionally, it is given that $ Y= X^2 $
We first make the following assumptions:
- $ f_{X_1}(x) $, and hence $ f_{X_2}(x) $ are even functions.
- the Fourier transforms of $ f_{X_1}(x) $ and $ f_{X_2}(x) $ exist.
then we want to show that
$ X_1, X_2 \sim N(\mu,\frac{1}{2}) \Leftrightarrow f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}} $
To prove sufficiency,
If $ X_1 $ and $ X_2 $ are two Gaussian distributed scalar random variables with mean $ \mu $ and variance $ \sigma^2 $,
$ E[e^{-jty}] = E[e^{-jtx^2}] = \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx $
We want to show that
$ X_1, X_2 \sim N(\mu,\frac{1}{2}) \Leftrightarrow f_Y(y) = \frac{e^{-\frac{y}{2}}}{\sqrt{2\pi y}} $
Let $ u = x^2 $, then
$ dx = \frac{du}{2x} $
and we have that
$ \begin{align} E[e^{-jtx^2}] &= \int_{-\infty}^{\infty}e^{-jtx^2}f_X(x)dx \\ &= \int_{-\infty}^{\infty}e^{-jtu}\frac{f_X(x)}{2x}du \\ &= \int_{-\infty}^{\infty}e^{-jtu}\frac{f_X(\sqrt{u})}{2\sqrt{u}}du \\ &= \mathcal{F}\{\frac{f_X(\sqrt{x})}{2\sqrt{x}}\} \end{align} $
Since $ Y= X^2 $, we have that
$ \begin{align} \mathcal{F}\{\frac{f_X(\sqrt{x})}{2\sqrt{x}}\} &= \mathcal{F}\{\frac{e^{-\frac{x}{2}}}{\sqrt{2\pi x}}\} \\ \Leftrightarrow \frac{f_X(\sqrt{x})}{2\sqrt{x}} &= \frac{e^{-\frac{x}{2}}}{\sqrt{2\pi x}} \\ \Leftrightarrow f_X(\sqrt{x}) &= \frac{\sqrt{2}e^{-\frac{x}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow f_X(x) &= \frac{\sqrt{2}e^{-\frac{x^2}{2}}}{\sqrt{\pi}} \end{align} $
Recall that $ X = X_1 - X_2 $, where $ X_1 $ and $ X_2 $ are iid. Therefore,
$ \begin{align} f_1(x) * f_2(x) &= \frac{\sqrt{2}e^{\frac{-x^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow f_1(x) * f_1(-x) &= \frac{\sqrt{2}e^{\frac{-x^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow F_1(t) . F_1(-t) &= \frac{\sqrt{2}e^{\frac{-t^2}{2}}}{\sqrt{\pi}} \\ \Leftrightarrow |F_1(t)|^2 &= \frac{\sqrt{2}e^{\frac{-t^2}{2}}}{\sqrt{\pi}} \end{align} $
where $ g(t) $ is the phase.
$ F_1(t) $ is real if we can assume that $ X_1 $ and hence $ X_2 $ are even functions. Then $ g(t) = 0 $ and $ F_1(t) $ is given by
$ F_1(t) = $
Taking the inverse Fourier transform, we get
$ f_1(x) = \frac{2e^{-x^2}}{\sqrt{\pi}} $
