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by KvanteKat
599 days ago
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For those interested in looking slightly more into the characteristic function, it may be worth pointing out that the characteristic function is equal to the Fourier-transform (with the sign of the argument being reversed) of the probability distribution in question. In my own experience teaching teaching probability theory to physicists and engineers, establishing this connection is often a good way of helping people build intuition for why characteristic functions are so useful, why they crop up everywhere in probability theory, and why we can extract so much useful information about a distribution by looking at the characteristic function (since this group of students tends to already be rather familiar with Fourier-transforms). |
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If you don't say that this is what you are doing then it all seems quite mysterious.