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LocalVolatility
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Consider a random variable $X$ withthat has a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus

\begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation}

Since $f(x) \geq 0$ for it to be a valid PDF, it follows that

\begin{equation} \int_0^\infty x f(x) \mathrm{d}x = 0 \qquad \Leftrightarrow \qquad f(x) = \delta(x), \end{equation}

where $\delta(x)$ is the Dirac delta function. I.e. for $X$ to have a zero-mean, its PDF needs to be a point mass as zero.

Consider a random variable $X$ with a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus

\begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation}

Since $f(x) \geq 0$ for it to be a valid PDF, it follows that

\begin{equation} \int_0^\infty x f(x) \mathrm{d}x = 0 \qquad \Leftrightarrow \qquad f(x) = \delta(x), \end{equation}

where $\delta(x)$ is the Dirac delta function. I.e. for $X$ to have a zero-mean, its PDF needs to be a point mass as zero.

Consider a random variable $X$ that has a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus

\begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation}

Since $f(x) \geq 0$ for it to be a valid PDF, it follows that

\begin{equation} \int_0^\infty x f(x) \mathrm{d}x = 0 \qquad \Leftrightarrow \qquad f(x) = \delta(x), \end{equation}

where $\delta(x)$ is the Dirac delta function. I.e. for $X$ to have a zero-mean, its PDF needs to be a point mass as zero.

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LocalVolatility
  • 6.1k
  • 4
  • 20
  • 35

Consider a random variable $X$ with a probability density function (PDF) $f(x)$. $X$ being non-negative means that $f(x) = 0$ for $x < 0$. The expectation of $X$ is thus

\begin{equation} \int_{-\infty}^\infty x f(x) \mathrm{d}x = \int_0^\infty x f(x) \mathrm{d}x. \end{equation}

Since $f(x) \geq 0$ for it to be a valid PDF, it follows that

\begin{equation} \int_0^\infty x f(x) \mathrm{d}x = 0 \qquad \Leftrightarrow \qquad f(x) = \delta(x), \end{equation}

where $\delta(x)$ is the Dirac delta function. I.e. for $X$ to have a zero-mean, its PDF needs to be a point mass as zero.