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I have random variables $(X_1, \dots, X_k)$ distributed independently according to normal distributions with complex means, i.e. $j\mu_i, i=1\dots k, j^2=-1$, with unit variances.

I want to study the random variable $$ Z = \sum_{i=1}^k X_i^2, $$ Can I use directly the result of the moment-generating function of Chi-squared distribution for $Z$? $$ M(t) = \frac{\exp(\frac{\lambda t }{ (1 - 2t)})}{(1-2t)^{k/2}}. $$ with $$\lambda = \sum_{i=1}^k j^2\mu_i^2 = - \sum_{i=1}^k \mu_i^2$$

kjetil b halvorsen
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user2843539
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  • Your title says "non-central Chi-squared distribution" while your text says "normal distributions". Both say "complex mean" while your text says "unit variances". So what is the distribution of each $X_i$? – Henry Feb 05 '20 at 08:29
  • Sorry for the confusion. I updated the question – user2843539 Feb 05 '20 at 08:45
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    That helps, but the next question is whether you really want $\sum X_i^2$ or $\sum \left|X_i^2\right|$ if $X_i$ is complex – Henry Feb 05 '20 at 08:57
  • In fact, the original problem is that I want to compute $\mathbb{E}[\exp(t(X - j\mu)^2)] $ where $X \sim \mathcal{N}(0,1)$. When following this question, https://stats.stackexchange.com/questions/168371/moment-generating-function-mgf-of-non-central-chi-squared-distribution, it seems like I can compute $\lambda$ like above – user2843539 Feb 05 '20 at 11:34
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    You have significantly overcomplicated your original problem: why don't you just ask about that directly? – whuber Feb 05 '20 at 15:53

1 Answers1

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According to a comment, it looks like your problem is to compute

$$E\left[\exp(t(X-j\mu)^2)\right]$$

where $X$ has a standard Normal distribution and, quite possibly, $j^2=-1.$ But that detail scarcely matters, because in any case $(X-j\mu)^2$ has (by definition) a non-central Chi-squared distribution and this expectation is its moment generating function,

$$\psi_{(X-j\mu)^2}(t) = \exp\left(i j\mu t\right)\left(1-2it\right)^{-1/2}$$

where $i^2=-1$ is the Complex unit. When $j=i$ this reduces to

$$\psi_{(X-i\mu)^2}(t) = \exp\left(-\mu t\right)\left(1-2it\right)^{-1/2}.$$

The justification for just plugging a complex value into the non-centrality parameter is that every step in the derivation of $\psi$ (by integrating against the standard Normal density function) goes through even with complex values: it just comes down to completing the square and noting that the resulting (exponential) integrand (a Gaussian with imaginary mean) is holomorphic in the region between the x-axis and any line parallel to it, thereby assuring the integral over that parallel line equals the corresponding real integral.

whuber
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