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Omitted exponent corrected. This must be what was meant.
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Douglas Zare
Douglas Zare
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Given $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$, for $i = 1,\dots,n$. How does one find the distribution of $D = \sum_{i=1}^n X_i$$D = \sum_{i=1}^n X_i^2$? In the case that all the standard deviations are the same (i.e. $\sigma_i = \sigma_j$$\sigma_i = \sigma_1$ for all $i=j$$i$), the random variable $D/\sigma_1^2$ has a noncentral chi-squared distribution. But when I don't have this simplifying assumption I don't know what to do. Is there a well known distribution for this?

Given $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$, for $i = 1,\dots,n$. How does one find the distribution of $D = \sum_{i=1}^n X_i$? In the case that all the standard deviations are the same (i.e. $\sigma_i = \sigma_j$ for all $i=j$), the random variable $D/\sigma_1^2$ has a noncentral chi-squared distribution. But when I don't have this simplifying assumption I don't know what to do. Is there a well known distribution for this?

Given $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$, for $i = 1,\dots,n$. How does one find the distribution of $D = \sum_{i=1}^n X_i^2$? In the case that all the standard deviations are the same (i.e. $\sigma_i = \sigma_1$ for all $i$), the random variable $D/\sigma_1^2$ has a noncentral chi-squared distribution. But when I don't have this simplifying assumption I don't know what to do. Is there a well known distribution for this?

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Jacqueline Nolis
Jacqueline Nolis
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Sum of Squares of Normal distributions

Given $X_i \sim \mathcal{N}(\mu_i,\sigma_i^2)$, for $i = 1,\dots,n$. How does one find the distribution of $D = \sum_{i=1}^n X_i$? In the case that all the standard deviations are the same (i.e. $\sigma_i = \sigma_j$ for all $i=j$), the random variable $D/\sigma_1^2$ has a noncentral chi-squared distribution. But when I don't have this simplifying assumption I don't know what to do. Is there a well known distribution for this?