Este es un post sólo para que una compañera de la universidad pueda checar. Espero que no los moleste.
241) \( f(x)=\displaystyle\frac{1}{2}\lambda e^{-\lambda \left |{x}\right |} \), \( x\in{\mathbb{R}} \) y \( \lambda>0 \)
Entonces
\( E[X]=\displaystyle\int_{-\infty}^{\infty}x\displaystyle\frac{1}{2}\lambda e^{-\lambda\left |{x}\right |}dx \)
\( =\displaystyle\frac{1}{2}\lambda(\displaystyle\int_{-\infty}^{0}xe^{\lambda x}dx+\displaystyle\int_{0}^{\infty}xe^{-\lambda x}dx) \)
\( =\displaystyle\frac{1}{2}\lambda((\displaystyle\frac{xe^{\lambda x}}{\lambda}-
\displaystyle\frac{e^{\lambda x}}{\lambda^2})|^{-\infty}_0+(\displaystyle\frac{xe^{-\lambda x}}{-\lambda}-\displaystyle\frac{e^{-\lambda x}}{\lambda^2})|^{0}_\infty) \)
\( =\displaystyle\frac{1}{2}\lambda(\displaystyle\frac{-1}{\lambda^2}+\displaystyle\frac{1}{\lambda^2})=0 \)
\( Var[X]=E[X^2]-E[X]^2=E[X^2] \) por ser \( E[X]=0 \)
Luego
\( Var[X]=\displaystyle\frac{1}{2}\lambda\displaystyle\int_{-\infty}^{\infty}x^2e^{-\lambda\left |{x}\right |}dx=\displaystyle\frac{1}{2}\lambda(\displaystyle\int_{-\infty}^{0}x^2e^{\lambda x}dx+\displaystyle\int_{0}^{\infty}x^2e^{-\lambda x}dx) \)
\( =\displaystyle\frac{1}{2}\lambda(\displaystyle\frac{x^2e^{\lambda x}}{\lambda}|^{-\infty}_0-\displaystyle\frac{2}{\lambda}\displaystyle\int_{-\infty}^{0}xe^{\lambda x}dx+\displaystyle\frac{x^2e^{-\lambda x}}{-\lambda}|^{0}_\infty-\displaystyle\frac{-2}{\lambda}\displaystyle\int_{0}^{\infty}xe^{-\lambda x}dx) \)
\( =\displaystyle\frac{1}{2}\lambda(-\displaystyle\frac{2}{\lambda}(-\displaystyle\frac{1}{\lambda^2})+\displaystyle\frac{2}{\lambda}(\displaystyle\frac{1}{\lambda^2}))=\displaystyle\frac{2}{\lambda^2} \)
246)
\( E[X]=\displaystyle\int_{0}^{\infty}x\displaystyle\frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}dx=\displaystyle\frac{\lambda^n}{\Gamma(n)}\displaystyle\int_{0}^{\infty}x^{n}e^{-\lambda x}dx \)
\( =\displaystyle\frac{\lambda^n}{\Gamma(n)}\displaystyle\int_{0}^{\infty}(\displaystyle\frac{u}{\lambda})^n e^{-u}\displaystyle\frac{du}{\lambda} \)
\( =\displaystyle\frac{1}{\lambda \Gamma(n)}\displaystyle\int_{0}^{\infty}u^n e^{-u}du=\displaystyle\frac{1}{\lambda \Gamma(n)}\Gamma(n+1)=\displaystyle\frac{n\Gamma(n)}{\lambda\Gamma(n)} \)
\( =\displaystyle\frac{n}{\lambda} \)
\( E[X^m]=\displaystyle\int_{0}^{\infty}x^m\displaystyle\frac{\lambda^n}{\Gamma(n)}x^{n-1}e^{-\lambda x}dx \)
\( =\displaystyle\frac{\lambda^n}{\Gamma(n)}\displaystyle\int_{0}^{\infty}x^{m+n-1}e^{-\lambda x}dx \)
\( =\displaystyle\frac{\lambda^n}{\Gamma(n)}\displaystyle\int_{0}^{\infty}(\displaystyle\frac{u}{\lambda})^{m+n-1}e^{-u}\displaystyle\frac{du}{\lambda} \)
\( =\displaystyle\frac{\lambda^n}{\Gamma(n)\lambda^{n+m}}\displaystyle\int_{0}^{\infty}u^{m+n-1}e^{-u}du \)
\( =\displaystyle\frac{1}{\Gamma(n)\lambda^m}\Gamma(m+n) \)
\( Var[X]=E[X^2]-E[X]^2=\displaystyle\frac{1}{\Gamma(n)\lambda^2}\Gamma(n+2)-\displaystyle\frac{n^2}{\lambda^2}=\displaystyle\frac{n^2+n-n^2}{\lambda^2}=\displaystyle\frac{n}{\lambda^2} \)
donde hemos usado que \( \Gamma(n+2)=(n+1)n\Gamma(n) \)
267) \( f(x)=\displaystyle\frac{1}{x}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(\ln(x)-\mu)^2}{2\sigma^2}) \), \( x>0 \)
\( E[X]=\displaystyle\int_{0}^{\infty}x\displaystyle\frac{1}{x}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(\ln(x)-\mu)^2}{2\sigma^2})dx \)
\( =\displaystyle\int_{0}^{\infty}\displaystyle\frac{1}{x}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(\ln(x)-\mu)^2}{2\sigma^2})dx \)
\( =\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}e^{-\displaystyle\frac{(y-\mu)^2}{2\sigma^2}}e^ydy \) donde hemos hecho \( \ln(x)=y \)
\( =\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}e^{-\displaystyle\frac{(y-(\mu+\sigma^2))^2}{2\sigma^2}}e^{\displaystyle\frac{\sigma^2}{2}+\mu}dy \)
\( =e^{\displaystyle\frac{\sigma^2}{2}+\mu}\underbrace{\displaystyle\int_{-\infty}^{\infty}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}e^{-\displaystyle\frac{(y-(\mu+\sigma^2))^2}{2\sigma^2}}dy}_{1} \) pues esa integral es la integral de una función de densidad
\( =e^{\displaystyle\frac{\sigma^2}{2}+\mu} \)
\( E[\ln X]=\displaystyle\int_{0}^{\infty}\ln(x)\displaystyle\frac{1}{x}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(\ln(x)-\mu)^2}{2\sigma^2}) \)
\( =\displaystyle\int_{-\infty}^{\infty}y\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(y-\mu)^2}{2\sigma^2})dy \) donde hemos hecho \( y=\ln(x) \)
\( =E[Y] \) pues como \( X \) sigue una distribución lognormal, entonces por definición, \( Y=\ln X \) es normal, y viendo la última integral (la que va antes de \( E[Y] \)), se tiene \( Y\sim{N(\mu,\sigma^2)} \) .
Luego, \( E[\ln X]=E[Y]=\mu \).
Finalmente
\( E[\ln ^2 X]=\displaystyle\int_{0}^{\infty}\ln ^2(x)\displaystyle\frac{1}{x}\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(\ln(x)-\mu)^2}{2\sigma^2}) \)
\( =\displaystyle\int_{-\infty}^{\infty}y^2\displaystyle\frac{1}{\sqrt[ ]{2\pi}\sigma}\exp(-\displaystyle\frac{(y-\mu)^2}{2\sigma^2})dy \)
\( =E[Y^2] \)...(*)
donde nuevamente como en el anterior, \( Y\sim{N(\mu,\sigma^2)} \).
Por otra parte, en general si \( Z \) es variable aleatoria, se tiene \( E[Z^2]=Var[Z]+E[Z]^2 \). En particular, en nuestro caso es
\( E[Y^2]=Var[Y]+E[Y]^2=\sigma^2+\mu^2 \).
Sustituyendo lo anterior en (*):
\( E[\ln ^2 X]=E[Y^2]=\sigma^2+\mu^2 \)
Por tanto, \( Var[\ln X]=E[\ln ^2 X]-E[\ln X]^2=\sigma^2+\mu^2-\mu^2=\sigma^2 \)
Espero que te sirva.
Saludos
