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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*}

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.

Covariance in normal lognormal (NLN) mixture

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

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.

Result and proof on the conditional expectation of the product of two random variables

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

conditional expectations for independent random variables

Rule with independent random variables and conditional expectations

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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This solution is due to the suggestions and corrections from @Hunaphu, @whuber. 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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  • $\begingroup$ This is of course frankly wrong. To treat the varying $c=-k/y$ as if it is constant is bound to provoke chaos. $\endgroup$ – Did Jul 17 '15 at 6:44
  • $\begingroup$ @Did could you please elaborate on your thoughts .. $\endgroup$ – texmex Jul 17 '15 at 12:23
  • $\begingroup$ I just did: constant $\ne$ random variable. $\endgroup$ – Did Jul 17 '15 at 14:00
  • $\begingroup$ @Did Please note I have changed the solution now. Do you still see any issues? $\endgroup$ – texmex Jul 22 '15 at 2:36
  • $\begingroup$ Sure--you seem to misapply the definition of conditional expectation. In your case, $$E(e^XY+k\mid e^XY+k>0)=\frac{E((e^XY+k)\,\mathbf 1_{e^XY+k>0})}{P(e^XY+k>0)}$$ but we see nowhere a computation of $$P(e^XY+k>0).$$ Instead you seem to compute something akin to $$\int \frac{E(e^xY+k;e^xY+k>0)}{P(e^xY+k>0)}f(x)dx,$$ anyway something unrelated to the quantity of interest. (And frankly, I did not check further down the solution.) $\endgroup$ – Did Jul 22 '15 at 7:35

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