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I need your help in solving the following problem: I have to calculate the characteristic function for the positive side of a random variable - simplest case: let $Y$ be standard normal and $Y^+ =\max (0,Y)$.

What is then the characteristic function of $Y^+$? I need to calculate both the direct formula (used as a check), but also the relation between $\phi_t(Y)$ and $\phi_t(Y^+)$ (to be used for more complex examples).

I have tried to get a straightforward answer by computing $\int_{-\infty}^{\infty}e^{ity^+}f(y)\mathrm dy$, by splitting in two integrals, completing the square and getting in the end: $\phi_t(Y) = \Phi(0) [ 1 - e^{\frac {-t^2} 2 } ]$ but I have no confidence in this result.

Any ideas on how this can be calculated?

Thank you for any feedback you have, Mihai

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There is no reason why $\varphi_Z(t)=\mathbb E(\mathrm e^{\mathrm itZ})$ with $Z=\max(Y,0)$ should be a simple function of $\varphi_Y(t)=\mathbb E(\mathrm e^{\mathrm itY})$, in general.

In the specific case when $Y$ is standard normal, the tedious residue computations one imagines yield $\varphi_Y(t)=\mathrm e^{-t^2/2}$ and $$ \varphi_Z(t)=\tfrac12+\mathrm e^{-t^2/2}\left(\tfrac12+\mathrm iG(t)\right),\qquad G(t)=\int_0^t\frac1{\sqrt{2\pi}}\mathrm e^{u^2/2}\mathrm du. $$ Edit: To prove this, write $\varphi_Z(t)=\mathbb P(Y\leqslant0)+\mathbb E(\mathrm e^{\mathrm itY};Y\gt0)=\frac12+\mathbb E(\mathrm e^{\mathrm itY};Y\gt0)$ and $$ \mathbb E(\mathrm e^{\mathrm itY};Y\gt0)=\int_0^{+\infty}\mathrm e^{\mathrm itx}\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}I(t), $$ with $$ I(t)=\int_0^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-(x-\mathrm it)^2/2}\mathrm dx. $$ Consider the rectangle with vertices $-\mathrm it$, $-\mathrm it+R$, $+R$ and $0$. The integral of the function $z\mapsto\mathrm e^{-z^2/2}$ along the boundary of this rectangle, considered clockwise, is zero, hence $$ I_R(t)-J_R(t)-I_R(0)+J_0(t)=0, $$ with $$ I_R(t)=\int_{-\mathrm it}^{R-\mathrm it}\frac1{\sqrt{2\pi}}\mathrm e^{-z^2/2}\mathrm dz, \quad J_R(t)=\int_R^{R-\mathrm it}\frac1{\sqrt{2\pi}}\mathrm e^{-z^2/2}\mathrm dz. $$ When $R\to+\infty$, $I_R(t)\to I(t)$ and $J_R(t)\to0$, hence $I(t)=I(0)-J_0(t)$ with $$ I(0)=\int_0^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\frac12, $$ and, by the change of variable $z=-\mathrm iu$, $$ J_0(t)=-\mathrm i\int_0^t\frac1{\sqrt{2\pi}}\mathrm e^{u^2/2}\mathrm du. $$ This completes the proof.

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  • $\begingroup$ the relationship between $\phi_z$ is in general expressed through a FFT inverse of $\phi_y$ and this is what i am trying to test: $\phi_z(u) = F_{0,v}^{-1}[ \frac {\phi_y(u+v)-\phi(v) } {iv} ] + 1$ $\endgroup$ – MihaiR Mar 22 '13 at 20:50
  • $\begingroup$ ,i can't reproduce your result for $\phi_z(t)$. so far i have: $\phi_z(t) = \int_{-\infty}^0 e^{\frac {-x^2} 2 }dx + \frac 1 {\sqrt{2\pi}} \int_0^\infty \frac {e^{(x-it)^{2}}} 2 dx = \frac 1 2 + I_1$, but i don't know how to calculate $I_1$ $\endgroup$ – MihaiR Mar 22 '13 at 21:09
  • $\begingroup$ There was a typo. Complete proof added. $\endgroup$ – Did Mar 22 '13 at 21:37
  • $\begingroup$ Thanks a lot Did. Still trying to match the results with the FFT version. $\endgroup$ – MihaiR Apr 3 '13 at 2:16
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An expression for the characteristic function of the random variable $\max(0,X)$ in terms of the Hilbert transform of the characteristic function of an arbitrary random variable $X$ is given in the note just posted at http://arxiv.org/abs/1309.5928 .

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