How do I go about finding the the pdf of the statisitc $\sum_i x_i ^2$ such that each $x_i$ is iid from a $N(\sigma , \sigma)$ distribution? I've searched, but cannot find a straightforward answer. I've looked at charts such as this (which I highly recommend reviewing if you have not seen one like it). I've also reviewed this answer about the noncentral chi-squared distribution, but it seems to be specifically for iid variables drawn from the distribution $N(0, \sigma^2)$.
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- $\begingroup$ Just to be clear: $X\sim N(\sigma,\sigma)$ means that $X$ has mean $\sigma$ and variance $\sigma$ as usual? $\endgroup$Jack D'Aurizio– Jack D'Aurizio2015-05-06 14:55:07 +00:00Commented May 6, 2015 at 14:55
- $\begingroup$ @JackD'Aurizio yes that is the case. $\endgroup$kathystehl– kathystehl2015-05-06 15:51:36 +00:00Commented May 6, 2015 at 15:51
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2 If $X_i$ are iid normal with mean $\mu$ and variance $\sigma^2$, then $\sigma^{-2} \sum_{i=1}^n X_i^2$ has noncentral chi-squared distribution with $n$ degrees of freedom and noncentrality parameter $\lambda = n (\mu/\sigma)^2$ (as defined e.g. in Wikipedia; conventions may sometimes differ). So what you have is a multiple of a noncentral chi-squared random variable. If $Y$ has pdf $f_Y$, then $Z = c Y$ (for $c > 0$) has pdf $f_Z(z) = c^{-1} f_Y(z/c)$.
- $\begingroup$ Thank you @Robert. To clarify, what would $c$ be in the example above? $\endgroup$kathystehl– kathystehl2015-05-06 15:54:57 +00:00Commented May 6, 2015 at 15:54
- $\begingroup$ Let me use $a$ instead of your $\sigma$, to avoid confusion. If $\mu = \sigma^2 = a$, then $Y = \sigma^{-2} \sum_{i=1}^n X_i^2$ has the noncentral chi-square distribution with $\lambda = n a$, and your statistic $Z = \sum_{i=1}^n X_i^2 = \sigma^2 Y = a Y$. $\endgroup$Robert Israel– Robert Israel2015-05-07 02:44:25 +00:00Commented May 7, 2015 at 2:44