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Could someone please provide the answer and steps to solve this expression?

\begin{eqnarray*} E\left[\left.\left(e^{X}Y+k\right)\right|\left.\left(e^{X}Y+k\right)>0\right]\right. \end{eqnarray*} $E$ is the expectation operator. \begin{eqnarray*} X\sim N\left(\mu_{X},\sigma_{X}^{2}\right);Y\sim N\left(\mu_{Y},\sigma_{Y}^{2}\right);X\;\text{and }Y\text{ are independent. Also, }k<0 \end{eqnarray*}

KEY MISSING LINK

The above expression depends on proving the below general expression or at least showing that it holds for the special case of Normal and Log-Normal distribution.

$E[UV\mid UV>c] = E[UE[V\mid UV>c]]$

Here, $U,V$ are independent random variables that could be discrete or continuos and follow any probability distribution. $E$ is the expectation operator. $c$ is a constant.

Of course, for our main question we require only the case where one of them is normally distributed and the other is log-normal? Say, $U$ is log-normal and $V$ is normal. Is the above identity true for this special case?

This is posed as a separate question due to its importance: Proof of Simplification of Conditional Expectation of Product of Random Variables

STEPS TRIED

0) JOINT CONDITIONAL DENSITY

I am having difficulty coming up with the conditional joint density to use in the above expectation. The joint density function of just $X$ and $Y$ is straight forward and follows from the standard density function for the bivariate normal case. How would we incorporate the conditional aspect into the joint density function?

1) NORMAL LOG-NORMAL MIXTURE PAPER BY YANG

(Link: http://repec.org/esAUSM04/up.21034.1077779387.pdf)

This paper has the first four central moments without proof (Equation 5 in above paper). If someone could provide these proofs, that might shed more light on the problem above.

The variables in the Yang paper are correlated, which is easy to apply to above; but they also have zero mean in the paper, which in our case does not apply directly, since we have non zero mean.

2)OTHER RELATED LINKS

a) Interesting question about an expectation involving a slightly modified form of the normal log-normal mixture. Though this lacks the conditional aspect, and hence needs some modification before it can be used for the problem above.

https://math.stackexchange.com/questions/1142841/covariance-in-normal-lognormal-nln-mixture

b) Another question on the normal log-normal mixture though this lacks a deeper discussion.

https://math.stackexchange.com/questions/159818/combination-of-a-normal-r-v-with-a-log-normal-one

c) Question on conditional expectation of product of independent random variables. It would be good to know which aspects from this are applicable in our case.

https://math.stackexchange.com/questions/544410/result-and-proof-on-the-conditional-expectation-of-the-product-of-two-random-var

d) Other interesting questions on conditional expectation of independent random variables.

https://math.stackexchange.com/questions/380866/conditional-expectations-for-independent-random-variables?rq=1

https://math.stackexchange.com/questions/55524/rule-with-independent-random-variables-and-conditional-expectations?rq=1

3) TAYLOR SERIES APPROXIMATIONS

Would it be possible to use taylor serious approximations here? I am little confused due to the conditional expectation and the normal log normal mixture? Any pointers on whether this is possible and how to proceed further or whether this is not applicable here would be great.

4) USING STANDARD NORMAL (SEEMS LIKE A DEAD END)

I know that if we can express this sum using the standard normal as below, there is a solution. Please advice on how to do this or other alternatives to solve the above would be helpful as well. This seems to be a dead end as confirmed by experts on this forum. But still keeping here if someone discovers a way to continue using this approach.

\begin{eqnarray*} W=\left(e^{X}Y+k\right)<=>\mu+\sigma Z\text{ where, }Z\sim N\left(0,1\right) \end{eqnarray*}

\begin{eqnarray*} \left[W\sim N\left(\mu,\sigma^{2}\right)\Rightarrow W=\mu+\sigma Z\;;\; W>0\Rightarrow Z>-\mu/\sigma\right] \end{eqnarray*} We then need to determine, $\mu\text{ and }\sigma$.

We have for every standard normal distribution, $Z$, and for every $u,$ $Pr\left[Z>\text-u\right]=Pr\left[Z<u\right]=\mathbf{\Phi}\left(u\right)$. Here, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively. \begin{eqnarray*} E\left[\left.Z\right|Z>-u\right] & = & \frac{1}{\mathbf{\Phi}\left(u\right)}\left[\int_{-u}^{\infty}t\phi\left(t\right)dt\right]\\ & = & \frac{1}{\mathbf{\Phi}\left(u\right)}\left[\left.-\phi\left(t\right)\right|_{-u}^{\infty}\right]=\frac{\phi\left(u\right)}{\mathbf{\Phi}\left(u\right)} \end{eqnarray*} Hence we have, \begin{eqnarray*} E\left[\left.Y\right|Y>0\right] & = & \mu+\sigma E\left[\left.Z\right|Z>\left(-\frac{\mu}{\sigma}\right)\right]\\ & = & \mu+\frac{\sigma\phi\left(\mu/\sigma\right)}{\mathbf{\Phi}\left(\mu/\sigma\right)} \end{eqnarray*} Setting, $\psi\left(u\right)=u+\phi\left(u\right)/\Phi\left(u\right)$, \begin{eqnarray*} E\left[\left.Y\right|Y>0\right]=\sigma\psi\left(\mu/\sigma\right) \end{eqnarray*}

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  • $\begingroup$ @gung, Please note above the steps I tried. Expressing the above as a standard normal variable with appropriate mean and variance has a solution. So the trick would be to convert the above to a suitable standard normal. If there are alternatives, please also suggest. Thanks for your help. $\endgroup$ – texmex Jun 21 '15 at 6:09
  • $\begingroup$ I just also found the Paper on Normal and Log Normal mixture by Yang, but it treats correlated variables and am not yet sure how to get the conditional expectation. Please advice on closed form solutions that might exist for the above. $\endgroup$ – texmex Jun 21 '15 at 7:01
  • $\begingroup$ What is the subscript $T$ in your first formula? $\endgroup$ – user765195 Jun 24 '15 at 5:22
  • $\begingroup$ Sorry, I should either have removed that or clarified earlier it is time .. I am taking the expectation at time T .. Please let me know if you need any further details ... Any suggestions on answers and steps would be appreciated? $\endgroup$ – texmex Jun 24 '15 at 8:33
  • 2
    $\begingroup$ Any reference that analyzes the error would do it. First, though, I recommend that you simplify the question by eliminating redundant parameters. It can be reduced to one about $e^XY$ where $X$ has a mean-zero Normal distribution and $Y$ has an arbitrary Normal distribution. That simplification will clarify the nature of the problem and perhaps help you determine whether a Taylor series approximation would be appropriate. $\endgroup$ – whuber Jul 1 '15 at 14:36
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What is the intended use of the result? That bears on what form of answer is needed, to include whether a stochastic (Monte Carlo) simulation approach might be adequate, And even the bigger picture matter of is this problem necessary to solve, and did someone come up with this problem as a way of solving a higher level problem, and there might be a better approach to the higher level problem which doesn't require this.

Here is a stochastic (Monte Carlo) simulation solution in MATLAB.

a = 1; b = 2; c = 3; d = 4; k = -1; % Made up values for illustrative purpose
n = 1e8; % Number of replications
mux = 10; sigmax = 4; sigmay = 7; % Made up values for illustrative purposes
X = mux + sigmax * randn(n,1); Y = sigmay * randn(n,1); Y1 = a + b + c + d * Y;
success_index = exp(X).*Y1 > 0; % replications in which condition is true
num_success = sum(success_index);
Cond_Sample = exp(X(success_index)) .* Y1(success_index) + k;
disp([num_success mean(Cond_Sample) std(Cond_Sample)/sqrt(num_success)])
1.0e+09 *
0.058475265000000   1.502775087443930   0.057342191058931
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  • $\begingroup$ Meant to add, I feel that this is a such a fundamental combination of normal and log normal, so there must be some solution if not approximation available. If there is not, given the rather simple nature of the problem, any further progress or breakthroughs would be helpful to many it seems. I would be happy to put a bounty on this as soon as I can, which is after I get five more points. $\endgroup$ – texmex Jul 2 '15 at 12:03
  • $\begingroup$ At this point, i think I've given enough freebies for the benefit of some billionaire hedge fund tycoon. They can pony up if they want more specialized help and advice. Or is this an academic project? $\endgroup$ – Mark L. Stone Jul 2 '15 at 12:07
  • $\begingroup$ This is an academic project. I am not billionaire hedge fund tycoon :-) If this were not academic, then a simulation would have killed it long back. Finding a better answer would be mainly in the interest of gaining knowledge and spreading it … Right? $\endgroup$ – texmex Jul 2 '15 at 12:14
  • $\begingroup$ Perhaps we should collaborate. You can send an invite to my LinkedIn account linked in my profile. And I wasn't thinking you were the billionaire hedge fund tycoon - more likely that you were an employee working for the benefit of the billionaire hedge fund tycoon :). $\endgroup$ – Mark L. Stone Jul 2 '15 at 12:14
  • $\begingroup$ Since I still don't really know the big picture, i can't assess what kind of solution, if any, is really needed to this. And so why is any of this Normal or Lognormal, and not some fatter tailed distributions? The simulation approach will still hold up if distributions change, dependencies are introduced, etc. $\endgroup$ – Mark L. Stone Jul 2 '15 at 12:17
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Comments:

  1. The joint density is given by multiplying the densities since they are idp. One variable is just a paramater to the other.
  2. $Ye^X$ is not normally distributed so approach (4) wont work.

  3. The expressions below might allow you to find some approximation. If not they are relatively easy to evaluate with a computer.

Let $X\sim\mathcal{N}(\mu, \sigma^2)$, denote $c = -k/y$, and $d(y)=\log(-k/y)$. Let $Z=e^X$, then $$E[YZ\mid B] = E[YE[Z\mid B]],$$ and, $$E[Z\mid Z>c] = e^{\mu+\sigma^2/2}\frac{P(X>d-\sigma^2)}{P(X>d)},$$ if $c>0$ and $E[Z] = e^{\mu+\sigma^2/2}$ otherwise. Thus, since $c>0$ only if $Y<0$, $$E[YZ\mid B] = \frac{1}{2}e^{\mu+\sigma^2/2}\bigg(E[Y\mathbb{1}(Y>0] + E\Big[Y\frac{P(X>d(Y)-\sigma^2)}{P(X>d(Y))}\mathbb{1}(Y<0)\Big]\bigg).$$

The first part is simply $$\int_0^\infty y f_Y(y)\,dy,$$ and the second $$\int_{-\infty}^0 y \frac{P(X>log(-\frac{k}{y})-\sigma^2)}{P(X>log(-\frac{k}{y}))}f_Y(y)\,dy.$$

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  • $\begingroup$ Thanks for the helpful answer. Here, $k<0$, but I suppose we just change the sign and everything follows. Also, could you please elaborate on the steps to get the below two results? Does the first one follow simply from independence or does it involve some other special properly? #1) $$E[YZ\mid B] = E[YE[Z\mid B]],$$ #2) $$E[Z\mid Z>c] = e^{\mu+\sigma^2/2}\frac{P(X>d-\sigma^2)}{P(X>d)},$$ $\endgroup$ – texmex Jul 8 '15 at 13:47
  • $\begingroup$ Could you also please explain how you are setting up the conditional density here? $$E[ZY\mid ZY\in B] = \int\int_B zy f_{ZY\mid B}\,dz\,dy = \int y \Big(\int_{-k/y} z f_Z\,dz\Big) f_Y\,dy$$ $\endgroup$ – texmex Jul 8 '15 at 14:47
  • $\begingroup$ There was a mistake in the two-dimensional integral. If you look at it now it should be clear whence $E[YZ\mid B]$ came from-otherwise let me know! The conditional expectation is a standard result found in e.g. the wikipedia article for lognormal. You find it by integrating e^X f_X with some constraints. $\endgroup$ – Hunaphu Jul 10 '15 at 12:02
  • $\begingroup$ Much appreciative of your clarifications. I see how we would get this, $$E[YZ\mid B] = E[YE[Z\mid B]],$$ But I found this link, while trying to understand this more and perhaps I am confusing something, but would the answer at this link not apply in the our case. math.stackexchange.com/questions/544410/… $\endgroup$ – texmex Jul 10 '15 at 12:35
  • $\begingroup$ We have $E[YE[Z\mid B]\neq E[Y]E[Z\mid B]$ $\endgroup$ – Hunaphu Jul 13 '15 at 8:40
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This solution is due to the suggestions and corrections from @Hunaphu, @whuber, and others. Could someone please verify if all the steps make sense?

ANSWER STEPS START

Using some notational shortcuts,

Consider, \begin{eqnarray*} E\left[\left.\left(e^{X}Y+k\right)\right|\left(e^{X}Y+k\right)>0\right] & = & E\left[k\left|\left(e^{X}Y+k\right)>0\right.\right]+E\left[\left(e^{X}Y\right)\left|\left(e^{X}Y+k\right)>0\right.\right] \end{eqnarray*} \begin{eqnarray*} & = & k+E\left[\left.\left(Ye^{X}\right)\right|\left.\left(Ye^{X}+k\right)>0\right]\right. \end{eqnarray*} \begin{eqnarray*} & = & k+\int\int ye^{x}f\left(\left.ye^{x}\right|\left\{ ye^{x}+k\right\} >0\right)dxdy \end{eqnarray*} Here, $f\left(w\right)$ is the probability density function for $w$, Here, $f\left(w\right)$ is the probability density function for $w$, \begin{eqnarray*} & = & k+\int\int ye^{x}\frac{f\left(ye^{x};\left\{ ye^{x}+k\right\} >0\right)}{f\left(\left\{ ye^{x}+k\right\} >0\right)}dxdy \end{eqnarray*} \begin{eqnarray*} \left[\text{We note that, }ye^{x}>-k>0\Rightarrow y>0\right] \end{eqnarray*} \begin{eqnarray*} & = & k+\int\int ye^{x}\frac{f\left(y\right)f\left(e^{x};\left\{ ye^{x}+k\right\} >0\right)}{f\left(\left\{ ye^{x}+k\right\} >0\right)}dxdy \end{eqnarray*} \begin{eqnarray*} & = & k+\int y\left[\int\frac{e^{x}f\left(e^{x};\left\{ e^{x}>-\frac{k}{y}\right\} \right)}{f\left(e^{x}>-\frac{k}{y}\right)}dx\right]f\left(y\right)dy \end{eqnarray*} \begin{eqnarray*} & = & k+\int y\left[\int e^{x}f\left(e^{x}\left|\left\{ e^{x}>-\frac{k}{y}\right\} \right.\right)dx\right]f\left(y\right)dy \end{eqnarray*} \begin{eqnarray*} & = & k+\int_{0}^{\left(y<-k\right)}y\left[\int e^{x}f\left(e^{x}\left|\left\{ e^{x}>1\right\} \right.\right)dx\right]f\left(y\right)dy+\int_{\left(y>-k\right)}^{\infty}y\left[\int e^{x}f\left(e^{x}\left|\left\{ e^{x}<1\right\} \right.\right)dx\right]f\left(y\right)dy \end{eqnarray*} \begin{eqnarray*} & = & k+\int_{0}^{\left(-k\right)}y\left[E\left(\left.W\right|W>c\right)\right]f\left(y\right)dy+\int_{\left(-k\right)}^{\infty}y\left[E\left(\left.W\right|W<c\right)\right]f\left(y\right)dy\quad;\;\text{here, }W=e^{X}\text{ and }c=1 \end{eqnarray*} Simplifying the inner expectations, \begin{eqnarray*} E\left(\left.W\right|W>c\right)=\frac{1}{P\left(e^{X}>c\right)}\int_{c}^{\infty}w\frac{1}{w\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{ln\left(w\right)-\mu_{X}}{\sigma_{X}}\right]^{2}}dw \end{eqnarray*} Put $t=ln\left(w\right)$, we have, $dw=e^{t}dt$ \begin{eqnarray*} E\left(\left.W\right|W>c\right)=\frac{1}{P\left(X>ln\left(c\right)\right)}\int_{ln\left(c\right)}^{\infty}\frac{e^{t}}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{t-\mu_{X}}{\sigma_{X}}\right)^{2}}dt \end{eqnarray*} \begin{eqnarray*} t-\frac{1}{2}\left(\frac{t-\mu_{X}}{\sigma_{X}}\right)^{2}=-\frac{1}{2\sigma_{X}^{2}}\left(t-\left(\mu_{X}+\sigma_{X}^{2}\right)\right)^{2}+\mu_{X}+\frac{\sigma_{X}^{2}}{2} \end{eqnarray*} \begin{eqnarray*} E\left(\left.W\right|W>c\right)=\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(\mu_{X}+\sigma_{X}Z>ln\left(c\right)\right)}\int_{ln\left(c\right)}^{\infty}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{t-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]^{2}}dt\quad;Z\sim N\left(0,1\right) \end{eqnarray*} Put $s=\left[\frac{t-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]$ and $b=\left[\frac{ln\left(c\right)-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]$ we have, $ds=\frac{dt}{\sigma_{X}}$ \begin{eqnarray*} E\left(\left.W\right|W>c\right)=\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(Z>\frac{ln\left(c\right)-\mu_{X}}{\sigma_{X}}\right)}\int_{b}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^{2}}ds \end{eqnarray*} \begin{eqnarray*} =\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(Z<\frac{-ln\left(c\right)+\mu_{X}}{\sigma_{X}}\right)}\left[\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^{2}}ds-\int_{-\infty}^{b}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^{2}}ds\right] \end{eqnarray*} \begin{eqnarray*} =\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(Z<\frac{-ln\left(c\right)+\mu_{X}}{\sigma_{X}}\right)}\left[1-\Phi\left(b\right)\right]\quad;\Phi\text{ is the standard normal CDF} \end{eqnarray*} \begin{eqnarray*} =\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{\Phi\left(\frac{-ln\left(c\right)+\mu_{X}}{\sigma_{X}}\right)}\left[\Phi\left(-b\right)\right] \end{eqnarray*} Similarly for the other case, \begin{eqnarray*} E\left(\left.W\right|W<c\right)=\frac{1}{P\left(e^{X}<c\right)}\int_{0}^{c}w\frac{1}{w\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{ln\left(w\right)-\mu_{X}}{\sigma_{X}}\right]^{2}}dw \end{eqnarray*} Put $t=ln\left(w\right)$, we have, $dw=e^{t}dt$ \begin{eqnarray*} E\left(\left.W\right|W<c\right)=\frac{1}{P\left(X<ln\left(c\right)\right)}\int_{-\infty}^{ln\left(c\right)}\frac{e^{t}}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{t-\mu_{X}}{\sigma_{X}}\right)^{2}}dt \end{eqnarray*} \begin{eqnarray*} t-\frac{1}{2}\left(\frac{t-\mu_{X}}{\sigma_{X}}\right)^{2}=-\frac{1}{2\sigma_{X}^{2}}\left(t-\left(\mu_{X}+\sigma_{X}^{2}\right)\right)^{2}+\mu_{X}+\frac{\sigma_{X}^{2}}{2} \end{eqnarray*} \begin{eqnarray*} E\left(\left.W\right|W<c\right)=\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(\mu_{X}+\sigma_{X}Z<ln\left(c\right)\right)}\int_{-\infty}^{ln\left(c\right)}\frac{1}{\sigma_{X}\sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{t-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]^{2}}dt\quad;Z\sim N\left(0,1\right) \end{eqnarray*} Put $s=\left[\frac{t-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]$ and $b=\left[\frac{ln\left(c\right)-\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right]$ we have, $ds=\frac{dt}{\sigma_{X}}$ \begin{eqnarray*} E\left(\left.W\right|W<c\right)=\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(Z<\frac{ln\left(c\right)-\mu_{X}}{\sigma_{X}}\right)}\int_{-\infty}^{b}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^{2}}ds \end{eqnarray*} \begin{eqnarray*} =\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{P\left(Z<\frac{ln\left(c\right)-\mu_{X}}{\sigma_{X}}\right)}\left[\Phi\left(b\right)\right]\quad;\Phi\text{ is the standard normal CDF} \end{eqnarray*} \begin{eqnarray*} =\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{\Phi\left(\frac{ln\left(c\right)-\mu_{X}}{\sigma_{X}}\right)}\left[\Phi\left(b\right)\right] \end{eqnarray*} Using the results for the inner expectations, \begin{eqnarray*} E\left[\left.\left(e^{X}Y+k\right)\right|\left(e^{X}Y+k\right)>0\right]=k+\int_{0}^{\left(-k\right)}y\left[\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{\Phi\left(\frac{-ln\left(c\right)+\mu_{X}}{\sigma_{X}}\right)}\left[\Phi\left(-b\right)\right]\right]f\left(y\right)dy+\int_{\left(-k\right)}^{\infty}y\left[\frac{e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}}{\Phi\left(\frac{ln\left(c\right)-\mu_{X}}{\sigma_{X}}\right)}\left[\Phi\left(b\right)\right]\right]f\left(y\right)dy \end{eqnarray*} \begin{eqnarray*} =k+e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}\left[\int_{0}^{\left(-k\right)}y\left\{ \frac{\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} f\left(y\right)dy+\int_{\left(-k\right)}^{\infty}y\left\{ \frac{\Phi\left(-\left[\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right]\right)}{\Phi\left(-\left[\frac{\mu_{X}}{\sigma_{X}}\right]\right)}\right\} f\left(y\right)dy\right] \end{eqnarray*} \begin{eqnarray*} =k+e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}\left[\int_{0}^{\left(-k\right)}y\left\{ \frac{\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} f\left(y\right)dy+\int_{\left(-k\right)}^{\infty}y\left\{ \frac{1-\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{1-\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} f\left(y\right)dy\right] \end{eqnarray*} \begin{eqnarray*} & = & k+e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}\left[\left\{ \frac{\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} \int_{-\frac{\mu_{Y}}{\sigma_{Y}}}^{-\left(\frac{k+\mu_{Y}}{\sigma_{Y}}\right)}\left(\mu_{Y}+\sigma_{Y}z\right)\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}dz\right.\\ & & +\left.\left\{ \frac{1-\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{1-\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} \int_{-\left(\frac{k+\mu_{Y}}{\sigma_{Y}}\right)}^{\infty}\left(\mu_{Y}+\sigma_{Y}z\right)\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}dz\right]\quad;Z\sim N\left(0,1\right) \end{eqnarray*} \begin{eqnarray*} & = & k+e^{\left(\mu_{X}+\frac{1}{2}\sigma_{X}^{2}\right)}\left[\left\{ \frac{\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} \left\{ \mu_{Y}\left[\Phi\left(-\left[\frac{k+\mu_{Y}}{\sigma_{Y}}\right]\right)-\Phi\left(-\frac{\mu_{Y}}{\sigma_{Y}}\right)\right]-\frac{\sigma_{Y}}{\sqrt{2\pi}}\left[e^{-\frac{1}{2}\left(\frac{k+\mu_{Y}}{\sigma_{Y}}\right)^{2}}-e^{-\frac{1}{2}\left(\frac{\mu_{Y}}{\sigma_{Y}}\right)^{2}}\right]\right\} \right.\\ & & +\left.\left\{ \frac{1-\Phi\left(\frac{\mu_{X}+\sigma_{X}^{2}}{\sigma_{X}}\right)}{1-\Phi\left(\frac{\mu_{X}}{\sigma_{X}}\right)}\right\} \left\{ \mu_{Y}\left[1-\Phi\left(-\left[\frac{k+\mu_{Y}}{\sigma_{Y}}\right]\right)\right]+\frac{\sigma_{Y}}{\sqrt{2\pi}}\left[e^{-\frac{1}{2}\left(\frac{k+\mu_{Y}}{\sigma_{Y}}\right)^{2}}\right]\right\} \right] \end{eqnarray*}

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  • $\begingroup$ Yeah, that or just do stochastic simulation. Who are these people sitting in academic judgment who don't allow simulation-based solutions? There are Professors at the most prestigious universities in the world, journals, and books dealing just with stochastic simulation. Hint: call it stochastic simulation, not Monte Carlo simulation - it sounds more sophisticated and less "vulgar". $\endgroup$ – Mark L. Stone Jul 13 '15 at 15:32
  • $\begingroup$ @MarkL.Stone You are funny and right about the stochastic simulation. As I mentioned, at this stage, trhe question is not to make any professor or hedge fund manager happy. It is knowledge for its own sake ... :-) $\endgroup$ – texmex Jul 14 '15 at 6:14

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