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Using LOTUS: $$\text{E}(Y) = \int_{-\infty}^{\infty}dx \Big( x^{6} \cdot \frac{1}{\sqrt{2 \pi}} \text{exp}(-\frac{1}{2}x^{2}) \Big)$$ I have two functions multiplied inside the integral, so use Integration by Parts: $$\int_{a}^{b} f g' = [fg]_{a}^{b} - \int_{a}^{b} f'g$$ $$f = x^{6} \rightarrow f' = 6x^{5}$$ $$g' = \frac{1}{\sqrt{2 \pi}} \text{exp}\Big(-\frac{1}{2} x^{2}\Big) \rightarrow g \approx \frac{1}{1+e^{-x}}$$ so $$\text{E}(Y) = \Big[\frac{x^{6}}{1+e^{-x}}\Big]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} dx \Big( \frac{6x^{5}}{1+e^{-x}}\Big)$$ ... and I'm stuck

A.L. Verminburger
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4 Answers4

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Given that the integrand is even in a symmetric interval, then:

\begin{equation} E(Y)=\sqrt{\frac{2}{\pi}}\int\limits_{0}^{+\infty} x^{6}e^{-\frac{x^{2}}{2}} \,dx \end{equation}

Let $s=\frac{x^{2}}{2}$, which implies that: $x=\sqrt{2s} \Leftrightarrow dx=\frac{1}{\sqrt{2}}s^{-\frac{1}{2}}\,ds$. The limits do not change. So now we have:

\begin{equation} E(Y)=\sqrt{\frac{2}{\pi}}\int\limits_{0}^{+\infty} 8s^{3}e^{-s} \frac{1}{\sqrt{2}}s^{-\frac{1}{2}}\,ds=\frac{8}{\sqrt{\pi}}\int\limits_{0}^{+\infty} s^{3}e^{-s} s^{-\frac{1}{2}}\,ds=\frac{8}{\sqrt{\pi}}\underbrace{\int\limits_{0}^{+\infty} s^{\frac{5}{2}}e^{-s}\,ds}_{\Gamma{\left(\frac{7}{2}\right)}} \end{equation}

Knowing that $\Gamma\left(\frac{7}{2}\right)=\displaystyle\frac{15\sqrt{\pi}}{8}$, then we can conclude that:

\begin{equation} E(Y)=\frac{8}{\sqrt{\pi}}\times\frac{15\sqrt{\pi}}{8}=15 \end{equation}

Luis Sierra
  • 484
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Hint Let \begin{equation} I_k = \int_{\mathbb R} x^{2 k} e^{-x^2/2} dx \end{equation} Use $x^{2(k+1)}e^{-x^2/2} = x^{2k+1}(x e^{-x^2/2})$ and deduce a relation between $I_{k+1}$ and $I_k$.

Gribouillis
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Deriving the pdf of $Y$ first: $$ F_Y(y) = \int_{-y^{\frac{1}{6}}}^{y^{\frac{1}{6}}} \varphi(x)dx = \Phi(y^{\frac{1}{6}}) - \Phi(-y^{\frac{1}{6}}) \\ f_Y(y)=\frac{2}{\sqrt{2 \pi}6}y^{-\frac{5}{6}}e^{-\frac{y^{\frac{1}{3}}}{2}}, y > 0\\ E[Y] = \frac{1}{3 \sqrt{2 \pi}}\int_{0}^{\infty}y^{\frac{1}{6}}e^{-\frac{y^{\frac{1}{3}}}{2}}dy = $$ To substitute variables, set $t = y^{\frac{7}{6}} \Leftrightarrow dt = \frac{7}{6}y^{\frac{1}{6}}dy$ $$ \mathbf{E}Y = \frac{2}{7 \sqrt{2 \pi}}\int_{0}^{\infty}\exp\left(-\frac{t^{\frac{2}{7}}}{2}\right)dt $$ This integrates to $$ \mathbf{E}Y = \frac{2}{7 \sqrt{2 \pi}} \cdot \frac{105 \sqrt{\pi}}{\sqrt{2}}=15 $$

Brian Tung
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Alex
  • 19,395
  • I'm not making the edit because I'm not entirely certain this is what you mean, but if the integrand in your last integral is $e^{-\frac{t^{2/7}}{2}}$, I think it's somewhat easier to read if the $2/7$ is written horizontally rather than vertically. (Possibly easier still if written as $\exp\left(-\frac{t^{2/7}}{2}\right)$.) – Brian Tung Jun 19 '20 at 17:34
  • @BrianTung thanks, this $is$ what I meant. This certainly improves readability. – Alex Jun 19 '20 at 17:45
  • I added larger parentheses as a further typography tweak: \left( and \right). Feel free to retain or not as you see fit. – Brian Tung Jun 19 '20 at 18:43
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Note that if $B$ is a Brownian motion, we have $B_1 \sim \mathcal N (0,1)$. The Ito formula yields $$B_1^6 = \int_0^1 6B_s^5 d B_s + \frac 1 2 \int_0^1 30B_s^4 d s$$ and $$B_t^4 = \int_0^t 4B_s^3dB_s + \frac 1 2\int_0^t 12 B_s^2 d s$$ Using that $\int_0^t 4B_s^3dB_s$ is a martingale (thus has mean zero),Fubini and that $B_s$ has variance $s$ we have $$\Bbb E[B_t^4] = \frac 1 2 \int_0^t 12 \Bbb E[B_s^2] ds = \int_0^t 6 s ds = 3t^2$$ Since $\int_0^1 6B_s^5 d B_s $ is a martingale as well we have $$\Bbb E[Y] = \Bbb E[ B_1^6 ] = 15 \int_0^1\Bbb E[B_s^4]ds = 15\int_0^1 3s^2ds = 15$$

Falrach
  • 4,242