please help me with this problem of double exponential distributionenter image description here

  • $\begingroup$ Hi sir, its in my assignment of stochastic calculus. $\endgroup$ – Jay Dec 15 '15 at 15:10
  • $\begingroup$ I am following Stochastic calculus for finance 2 by steven shreve $\endgroup$ – Jay Dec 15 '15 at 15:29

Since $f_X(x) > 0$ and \begin{align*} \int_{-\infty}^{\infty} f_X(x) dx = 1, \end{align*} $f_X(x)$ is a valid density function.

Let \begin{align*} \varphi(x) =\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \end{align*} be the density function of a standard normal random variable. We define the measure $\widetilde{P}$ using the Randon-Nykodim derivative \begin{align*} \frac{d\widetilde{P}}{dP} = \frac{\varphi(X)}{f_X(X)}. \end{align*} Then, \begin{align*} \widetilde{P}(X \leq x) &= E_{\widetilde{P}}(1_{X \le x})\\ &=E_P\left(\frac{d\widetilde{P}}{dP} 1_{X \le x} \right)\\ &=E_P\left(\frac{\varphi(X)}{f_X(X)} 1_{X \le x} \right)\\ &=\int_{-\infty}^x \frac{\varphi(x)}{f_X(x)} f_X(x) dx\\ &=\int_{-\infty}^x \varphi(x) dx. \end{align*} That is, $X$ is standard normal w.r.t. the measure $\widetilde{P}$.

The equivalence follows, since \begin{align*} \frac{dP}{d\widetilde{P}} = \frac{f_X(X)}{\varphi(X)}. \end{align*}

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  • $\begingroup$ Thank you so much sir! Its really helpful in understanding the concept. $\endgroup$ – Jay Dec 15 '15 at 21:13

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