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Define $K_h$ as$$ K_h(x) = \dfrac{1}{h}K\bigg(\dfrac{x}{h}\bigg), $$ where $K$ is the Gaussian function$$ K(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}. $$

Define the function $T(x,y)$ by$$ T(x,y) = ((K_h*K_g - K_g)*(K_h*K_g - K_g))(x-y). $$

This expression arises in the context of bootstrapping for kernel density estimation. How can this expression be evaluated in Mathematica?

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    $\begingroup$ Try: k[x_] = 1/Sqrt[2 Pi] Exp[-x^2/2]; kh[x_] = 1/h K[x/h]; kg[x_] = 1/g K[x/g]; T[x_, y_] = ((kh[x] kg[x] - kg[x]) (kf[x] kg[x] - kg[x])) (x - y); T[x, y] $\endgroup$ – Daniel Huber Apr 6 at 16:18
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Clear["Global`*"]

k[h_, x_] = PDF[NormalDistribution[0, h], x]

(* E^(-(x^2/(2 h^2)))/(h Sqrt[2 π]) *)

Assuming that the first part takes the argument x - y rather than being distributed across x - y

t[x_, y_] = ((k[h, #]*k[g, #] - k[g, #])^2) &[x - y] // Simplify

(* (E^(-(((g^2 + h^2) (x - y)^2)/(
  g^2 h^2))) (-1 + 
   E^((x - y)^2/(2 h^2)) h Sqrt[2 π])^2)/(4 g^2 h^2 π^2) *)
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