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| -rw-r--r-- | hw1/main.tex | 24 |
1 files changed, 12 insertions, 12 deletions
diff --git a/hw1/main.tex b/hw1/main.tex index cd5c1f9..e05594e 100644 --- a/hw1/main.tex +++ b/hw1/main.tex @@ -60,7 +60,7 @@ $X+Y$ respectively, we have: g(z) = \int_{\mathbb{R}} f_X(u)f_Y(z-u)du,\quad z\in\mathbb{R} \end{displaymath} -(b) Using the formula for the cdf of a gaussain, we obtain: +(b) Using the formula for the cdf of a Gaussian variable, we obtain: \begin{displaymath} g(z) = \frac{1}{2\pi\sigma_X\sigma_Y}\int_{\mathbb{R}} \exp\left(-\frac{u^2}{2\sigma_X^2} - \frac{(z-u)^2}{2\sigma_y^2}\right)du @@ -78,8 +78,8 @@ Hence: g(z) = \frac{1}{2\pi\sigma_X\sigma_Y} \exp\left(- \frac{z^2}{2(\sigma_X^2+\sigma_Y^2)}\right) \int_{\mathbb{R}} - -\frac{\sigma_X^2+ \sigma_Y^2}{2\sigma_X^2\sigma_Y^2}\left(u - - \frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}z\right)^2du + \exp\left(-\frac{\sigma_X^2+ \sigma_Y^2}{2\sigma_X^2\sigma_Y^2}\left(u + - \frac{\sigma_X^2}{\sigma_X^2+\sigma_Y^2}z\right)^2\right)du \end{displaymath} We now recognize the integral of the un-normalized cdf of a Gaussian variable. Hence the integral evaluates to: @@ -91,12 +91,12 @@ after simplification we obtain the following formula for $g(z)$: g(z) = \frac{1}{\sqrt{2\pi}\sqrt{\sigma_X^2+\sigma_Y^2}} \exp\left(- \frac{z^2}{2(\sigma_X^2+\sigma_Y^2)}\right) \end{displaymath} -That is, $X+Y$ is a Gaussian variable of mean $0$ and variance +That is, $X+Y$ is a Gaussian variable with mean $0$ and variance $\sigma_X^2+\sigma_Y^2$. -(c) If $X$ has mean $\mu_X$ and $Y$ has mean $\mu_Y$, $X-\mu_X$ and $Y-\mu_Y$ -are still independent variables with the same variance as before and zero mean. -Applying the previous result. $X+Y-\mu_X-\mu_Y$ is a Gaussian variable with +(c) If $X$ has mean $\mu_X$ and $Y$ has mean $\mu_Y$, $X-\mu_X$ and $Y-\mu_Y$ +are still independent variables with unchanged variance and zero mean. +Applying the previous result, $X+Y-\mu_X-\mu_Y$ is a Gaussian variable with mean $0$ and variance $\sigma_X^2 + \sigma_Y^2$. Hence, $X+Y$ is normally distributed with mean $\mu_X+\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2$. @@ -106,16 +106,16 @@ for any interval $[a,b]$ with $a,b>0$: \begin{align*} \prob(Y\in[a,b]) = \prob(X\in [\log a, \log b]) &=\frac{1}{\sqrt{2\pi}\sigma_X}\int_{\log a}^{\log b} - \exp\left(-\frac{1}{2\sigma_X}(x-\mu)^2\right)dx\\ - &=\frac{1}{\sqrt{2\pi}\sigma_X}\int_{a}^{b} - \exp\left(-\frac{1}{2z\sigma_X}(\log z-\mu)^2\right)dz + \exp\left(-\frac{1}{2\sigma_X^2}(x-\mu)^2\right)dx\\ + &=\frac{1}{\sqrt{2\pi}\sigma_X^2}\int_{a}^{b} + \exp\left(-\frac{1}{2z\sigma_X^2}(\log z-\mu)^2\right)dz \end{align*} where we used the change of variable $z = e^x$ in the last equality. Hence the pdf of $Y$: \begin{displaymath} - f_Y = \begin{cases} + f_Y(z) = \begin{cases} \frac{1}{\sqrt{2\pi}\sigma_X} - \exp\left(-\frac{1}{2z\sigma_X}(\log z-\mu)^2\right)&\text{if $x>0$}\\ + \exp\left(-\frac{1}{2z\sigma_X^2}(\log z-\mu)^2\right)&\text{if $z>0$}\\ 0&\text{otherwise} \end{cases} \end{displaymath} |