# QUESTION

I am looking to find a simplification of the expression below. I have attempted this using the Taylor series. The question then remains if we can show the Lagrange remainder goes to zero. I have shown the steps to which I have reached. Could someone please verify / provide additional pointers / alternative solutions?

\begin{eqnarray*} \beta\left(x\right) & = & \left[x-\int_{0}^{x}\left[\frac{\Phi\left(\frac{lny-\mu}{\sigma}\right)}{\Phi\left(\frac{lnx-\mu}{\sigma}\right)}\right]^{M-1}dy\right] \end{eqnarray*}

Here, $\Phi(u)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{u}e^{-t^{2}/2}dt$ , the standard normal cumulative distribution and $X=e^{W}$where, $W\sim N\left(\mu,\sigma\right)$. $M$ is a non-zero positive integer. $\beta(x)=0$, when $x=0$

# STEPS TRIED

\begin{eqnarray*} \beta\left(x\right) & = & \left[x-\frac{\int_{0}^{x}\left[\Phi\left(\frac{lny-\mu}{\sigma}\right)\right]^{M-1}dy}{\left[\Phi\left(\frac{lnx-\mu}{\sigma}\right)\right]^{M-1}}\right]\\ & = & \left[x-\frac{\left\{ \int_{0}^{x}\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\left(\frac{lny-\mu}{\sigma}\right)}e^{-t^{2}/2}dt\right]^{M-1}dy\right\} }{\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\left(\frac{lnx-\mu}{\sigma}\right)}e^{-t^{2}/2}dt\right]^{M-1}}\right] \end{eqnarray*} Let, \begin{eqnarray*} h\left(y\right) & = & \left[\int_{-\infty}^{\left(\frac{lny-\mu}{\sigma}\right)}e^{-t^{2}/2}dt\right]^{M-1}\\ j\left(y\right) & = & \int h\left(y\right)dy \end{eqnarray*} We then have, \begin{eqnarray*} \beta\left(x\right) & = & \left[x-\frac{\left\{ \int_{0}^{x}h\left(y\right)dy\right\} }{h\left(x\right)}\right]\\ & = & \left[x-\frac{\left|j\left(y\right)\right|_{0}^{x}}{h\left(x\right)}\right]=\left[x-\left\{ \frac{j\left(x\right)-j\left(0\right)}{h\left(x\right)}\right\} \right]\\ & \approx & \left[x-\frac{j'\left(0\right)x}{h\left(x\right)}\right]\;\left\{ \because j\left(x\right)-j\left(0\right)\simeq j'\left(0\right)x\;,\quad Maclaurin\; Series\right\} \\ & = & \left[x-\frac{h\left(0\right)x}{h\left(x\right)}\right]=x\left[1-\frac{h\left(0\right)}{h\left(x\right)}\right] \end{eqnarray*} \begin{eqnarray*} \Rightarrow\beta\left(x\right) & = & x\;\left[\;\because h\left(0\right)=\left[\int_{-\infty}^{\left(\frac{ln0-\mu}{\sigma}\right)}e^{-t^{2}/2}dt\right]^{M-1}=\left[\int_{-\infty}^{-\infty}e^{-t^{2}/2}dt\right]^{M-1}=0\right] \end{eqnarray*} We could include additional terms, for greater precision, using the subsequent terms of the Maclaurin series, as follows, \begin{eqnarray*} \beta\left(x\right) & \approx & x\left[1-\frac{h\left(0\right)}{h\left(x\right)}-\frac{x}{2}\frac{h'\left(0\right)}{h\left(x\right)}\right]\;\left\{ \because j\left(x\right)-j\left(0\right)\simeq j'\left(0\right)x+\frac{j''\left(0\right)x^{2}}{2!}\right\} \\ \beta\left(x\right) & = & x\left[1-\frac{h\left(0\right)}{h\left(x\right)}-\frac{1}{2}\left\{ 1-\frac{h\left(0\right)}{h\left(x\right)}\right\} \right]\;\left\{ \because\frac{h\left(x\right)-h\left(0\right)}{h\left(x\right)}\simeq\frac{h'\left(0\right)x}{h\left(x\right)}\right\} \\ \Rightarrow\beta\left(x\right) & = & \frac{x}{2}\;\left[\;\because h\left(0\right)=0\right] \end{eqnarray*} \begin{eqnarray*} \beta\left(x\right) & \approx & x\left[1-\frac{h\left(0\right)}{h\left(x\right)}-\frac{x}{2}\frac{h'\left(0\right)}{h\left(x\right)}-\frac{x^{2}}{6}\frac{h''\left(0\right)}{h\left(x\right)}\right]\\ & & \;\left\{ \because j\left(x\right)-j\left(0\right)\simeq j'\left(0\right)x+\frac{j''\left(0\right)x^{2}}{2!}+\frac{j'''\left(0\right)x^{3}}{3!}\right\} \end{eqnarray*} \begin{eqnarray*} \beta\left(x\right) & \approx & x\left[1-\frac{h\left(0\right)}{h\left(x\right)}-\frac{x}{2}\frac{h'\left(0\right)}{h\left(x\right)}-\frac{1}{3}\left\{ 1-\frac{h\left(0\right)}{h\left(x\right)}-x\frac{h'\left(0\right)}{h\left(x\right)}\right\} \right]\\ & & \;\left\{ \because\frac{h\left(x\right)-h\left(0\right)-h'\left(0\right)x}{h\left(x\right)}\simeq\frac{1}{2}\frac{h''\left(0\right)x^{2}}{h\left(x\right)}\right\} \end{eqnarray*} \begin{eqnarray*} \beta\left(x\right) & \approx & x\left[1-\frac{h\left(0\right)}{h\left(x\right)}-\frac{x}{2}\frac{h'\left(0\right)}{h\left(x\right)}-\frac{1}{3}+\frac{1}{3}\frac{h\left(0\right)}{h\left(x\right)}+\frac{x}{3}\frac{h'\left(0\right)}{h\left(x\right)}\right]\\ & = & x\left[\frac{2}{3}-\frac{2}{3}\frac{h\left(0\right)}{h\left(x\right)}-\frac{1}{6}\left\{ 1-\frac{h\left(0\right)}{h\left(x\right)}\right\} \right]\\ \Rightarrow\beta\left(x\right) & = & \frac{x}{2}\;\left[\;\because h\left(0\right)=0\right] \end{eqnarray*}

Checking the Lagrange remainders $R^{M}\left(y\right)$ for a degree $M$ approximation where $0<\xi_{M}<y$, \begin{eqnarray*} R^{M}\left(y\right) & = & \frac{j^{M+1}\left(\xi_{M}\right)y^{M+1}}{\left(M+1\right)!}=\left|\frac{\partial^{M+1}j\left(y\right)}{\partial y^{M+1}}\right|_{y=\xi_{M}}\left[\frac{y^{M+1}}{\left(M+1\right)!}\right]\\ & = & \left|\frac{\partial^{M+1}\left\{ \int\left[\int_{-\infty}^{\left(\frac{lny-\mu}{\sigma}\right)}e^{-t^{2}/2}dt\right]^{M-1}\right\} }{\partial y^{M+1}}\right|_{y=\xi_{M}}\left[\frac{y^{M+1}}{\left(M+1\right)!}\right] \end{eqnarray*} Let, \begin{eqnarray*} l\left(y\right) & = & \left(\frac{lny-\mu}{\sigma}\right)\Rightarrow l'\left(y\right)=\frac{1}{\sigma y}\Rightarrow l''\left(y\right)=-\frac{1}{\sigma y^{2}} \end{eqnarray*} \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{M+1}}{\left(M+1\right)!}\right]\left|\frac{\partial^{M}\left\{ \left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{M-1}\right\} }{\partial y^{M}}\right|_{y=\xi_{M}} \end{eqnarray*}

Using Fa'adi Bruno's Formula, \begin{eqnarray*} \frac{\partial^{M}\left\{ p\left(q\left(y\right)\right)\right\} }{\partial y^{M}}=\sum\frac{M!}{k_{1}!\; k_{2}!\;...\; k_{M}!}p^{\left(k\right)}\left(q\left(y\right)\right)\left(\frac{q'\left(y\right)}{1!}\right)^{k_{1}}\left(\frac{q''\left(y\right)}{2!}\right)^{k_{2}}...\left(\frac{q^{\left(M\right)}\left(y\right)}{M!}\right)^{k_{M}} \end{eqnarray*} where the sum is over all nonnegative integer solutions of the Diophantine equation $k_{1}+2k_{2}+\cdot\cdot\cdot+Mk_{M}=M,\text{ and }k=k_{1}+k_{2}+\cdot\cdot\cdot+k_{M}$. Let, \begin{eqnarray*} p\left(y\right)=y^{M-1}\quad;q\left(y\right)=\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt \end{eqnarray*} \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{M+1}}{\left(M+1\right)!}\right]\\ & & \left|\sum\frac{M!}{k_{1}!\; k_{2}!\;...\; k_{M}!}p^{\left(k\right)}\left(q\left(y\right)\right)\left(\frac{q'\left(y\right)}{1!}\right)^{k_{1}}\left(\frac{q''\left(y\right)}{2!}\right)^{k_{2}}...\left(\frac{q^{\left(M\right)}\left(y\right)}{M!}\right)^{k_{M}}\right|_{y=\xi_{M}} \end{eqnarray*} $\text{As an illustration, let }k_{2}=\frac{M}{2}\Rightarrow k=\frac{M}{2}\text{ and let }K=\frac{M}{2}$ with $\triangle_{M}$ being the sum of the other terms, where each of the other terms is smaller than the term we consider. \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{2K+1}}{\left(2K+1\right)!}\right]\left|\frac{2K!}{K!}p^{\left(K\right)}\left(q\left(y\right)\right)\left(\frac{q''\left(y\right)}{2!}\right)^{K}+\triangle_{M}\right|_{y=\xi_{M}} \end{eqnarray*} \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{2K+1}}{\left(2K+1\right)!}\right]\\ & & \left|\frac{\left(2K!\right)\left(M-1\right)\;...\;\left(M-K\right)}{K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\left(e^{-\frac{1}{2}l\left(y\right)^{2}}\right)l'\left(y\right)\left[-\frac{1}{y}-l\left(y\right)l'\left(y\right)\right]}{2!}\right)^{K}+\triangle_{M}\right|_{y=\xi_{M}} \end{eqnarray*} \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{2K+1}}{\left(2K+1\right)!}\right]\\ & & \left|\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}y^{2}\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)2!}\right)^{K}+\triangle_{M}\right|_{y=\xi_{M}} \end{eqnarray*} Evaluating at the maximum value of $y$. \begin{eqnarray*} R^{M}\left(y\right) & = & \left[\frac{y^{2K+1}}{y^{2K}\left(2K+1\right)!}\right]\\ & & \left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M}\right] \end{eqnarray*} The next remainder term would be, \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{M+2}}{\left(M+2\right)!}\right]\left|\frac{\partial^{M+1}\left\{ \left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{M}\right\} }{\partial y^{M+1}}\right|_{y=\xi_{M}} \end{eqnarray*} \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{M+2}}{\left(M+2\right)!}\right]\\ & & \left|\sum\frac{\left(M+1\right)!}{k_{1}!\; k_{2}!\;...\; k_{M}!}p^{\left(k\right)}\left(q\left(y\right)\right)\left(\frac{q'\left(y\right)}{1!}\right)^{k_{1}}\left(\frac{q''\left(y\right)}{2!}\right)^{k_{2}}...\left(\frac{q^{\left(M+1\right)}\left(y\right)}{\left(M+1\right)!}\right)^{k_{M}}\right|_{y=\xi_{M+1}} \end{eqnarray*} In this case, $k_{1}+2k_{2}+\cdot\cdot\cdot+\left(M+1\right)k_{M+1}=M+1,\text{ and }k=k_{1}+k_{2}+\cdot\cdot\cdot+k_{M+1}$. $\;\text{As an illustration, let }k_{1}=1;\; k_{2}=\frac{M}{2}\Rightarrow k=\frac{M}{2}+1\text{ and let }K=\frac{M}{2}$ with $\triangle_{M+1}$ being the sum of the other terms, where each of the other terms is smaller than the term we consider. \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{2K+2}}{\left(2K+2\right)!}\right]\left|\frac{\left(2K+1\right)!}{1!K!}p^{\left(K+1\right)}\left(q\left(y\right)\right)\left(\frac{q'\left(y\right)}{1!}\right)^{1}\left(\frac{q''\left(y\right)}{2!}\right)^{K}+\triangle_{M+1}\right|_{y=\xi_{M}} \end{eqnarray*} \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{2K+2}}{\left(2K+2\right)!}\right]\left|\frac{\left(2K!\right)\left(M-1\right)\;...\;\left(M-K\right)\left(M-K-1\right)}{K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-2}\right. \end{eqnarray*} \begin{eqnarray*} \left.\left\{ \left(e^{-\frac{1}{2}l\left(y\right)^{2}}\right)l'\left(y\right)\right\} \left\{ \frac{\left(e^{-\frac{1}{2}l\left(y\right)^{2}}\right)l'\left(y\right)\left[-\frac{1}{y}-l\left(y\right)l'\left(y\right)\right]}{2!}\right\} ^{K}+\triangle_{M+1}\right|_{y=\xi_{M}} \end{eqnarray*} \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{2K+2}}{\left(2K+2\right)!}\right] \end{eqnarray*} \begin{eqnarray*} \left|\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)\left(K-1\right)}{K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-2}\left\{ \frac{1}{\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)\sigma y}\right\} \left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}y^{2}\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)2!}\right)^{K}+\triangle_{M+1}\right|_{y=\xi_{M}} \end{eqnarray*} Evaluating at the maximum value of $y$. \begin{eqnarray*} R^{M+1}\left(y\right) & = & \left[\frac{y^{2K+2}}{y^{2K+1}\left(2K+2\right)!}\right]\\ & & \left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)\left(K-1\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-2}\left\{ \frac{1}{\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)\sigma}\right\} \left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M+1}\right] \end{eqnarray*} Let us consider the ratio of two successive remainders. \begin{eqnarray*} \frac{R^{M+1}\left(y\right)}{R^{M}\left(y\right)} & = & \frac{\left[\frac{y^{2K+2}}{y^{2K+1}\left(2K+2\right)!}\right]\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)\left(K-1\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-2}\left\{ \frac{1}{\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)\sigma}\right\} \left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M+1}\right]}{\left[\frac{y^{2K+1}}{y^{2K}\left(2K+1\right)!}\right]\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M}\right]} \end{eqnarray*} \begin{eqnarray*} \frac{R^{M+1}\left(y\right)}{R^{M}\left(y\right)} & = & \frac{\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)\left(K-1\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-2}\left\{ \frac{1}{\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)\sigma}\right\} \left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}\right]}{\left(2K+2\right)\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M}\right]}\\ & & +\frac{\left[\triangle_{M+1}\right]}{\left(2K+2\right)\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M}\right]} \end{eqnarray*} \begin{eqnarray*} \frac{R^{M+1}\left(y\right)}{R^{M}\left(y\right)} & = & \frac{\left[\frac{\left(K-1\right)}{\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]\left(e^{\frac{1}{2}l\left(y\right)^{2}}\right)\sigma}\right]}{\left(2K+2\right)\left[1+\triangle_{M}\left\{ \frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}\right\} ^{-1}\right]}\\ & & +\frac{\left[\triangle_{M+1}\right]}{\left(2K+2\right)\left[\frac{\left(2K!\right)\left(2K-1\right)\;...\;\left(K\right)}{2^{K}K!}\left[\int_{-\infty}^{l\left(y\right)}e^{-t^{2}/2}dt\right]^{K-1}\left(\frac{\mu-\sigma^{2}-lny}{\sigma^{3}e^{\frac{1}{2}l\left(y\right)^{2}}}\right)^{K}+\triangle_{M}\right]} \end{eqnarray*}

• Unless $M$ has some very special value, you haven't a prayer of expanding this expression around $x=0$ because it's not even defined for negative $x$. What problem are you trying to solve that led to this exercise? – whuber Dec 22 '15 at 15:16
• Thanks whuber for checking this. You are right. This would be only for positive values … $$Beta(x)$$ is an auction Bid function using a log normal distribution function. Please let me know if you need any further details. M is from the set of positive integers – texmex Dec 22 '15 at 15:21
• Since it's not defined in a neighborhood of zero, it won't be differentiable there, either, so the Taylor series doesn't exist. The entire approach looks hopeless. Are you trying to approximate this function? If so, why not simply ask about ways to approximate it? – whuber Dec 22 '15 at 18:10
• Thanks @whuber ... The simplication would be the main goal. Perhaps I should modify the question to highlight the part before STEPS TRIED that I currently have. Please suggest if that would help. Also, you make a good point about the region of validity and upon further reflection. $x>0$ would be a valid assumption in this case. – texmex Dec 23 '15 at 3:51