How do we simplify this integral?

\begin{eqnarray*} \int_{-\infty}^{\infty}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy \end{eqnarray*}

Please note $k<0$ here. \begin{eqnarray*} Y\sim N\left(\mu_{Y},\sigma_{Y}^{2}\right); k<0 \end{eqnarray*}

Here, $f\left(y\right)$ is the probability density function for $y$, and $\mathbf{\Phi}$ is the standard normal CDF.


Based on other suggestions, please see related link below. It seems one of the two assertions below are valid. But I am not sure if (and which) of these are correct or how we can prove it? Could someone please clarify and provide steps?

I think the second assertion below holds when $\lim_{y\to0^+}$ though I am not sure and hence would appreciate clarifications as well.

How about other cases? (Can this integral be simplied in some region?)


\begin{eqnarray*} \left[\int_{-\infty}^{0}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy=\int_{-\infty}^{0}yf\left(y\right)dy\right] \end{eqnarray*}


\begin{eqnarray*} \left[\int_{0}^{\infty}\left\{ \frac{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\left(\mu_{X}+\sigma_{X}^{2}\right)}{\sigma_{X}}\right)}{\Phi\left(\frac{-ln\left(-\frac{k}{y}\right)+\mu_{X}}{\sigma_{X}}\right)}\right\} yf\left(y\right)dy=\int_{0}^{\infty}yf\left(y\right)dy\right] \end{eqnarray*}

This comes up during the proof for this question. Conditional Expected Value of Product of Normal and Log-Normal Distribution

  • $\begingroup$ I think it is impossible. ratio in the braces obviously not equal to 1, except when $\sigma_x^2$ equal to zero. $\endgroup$
    – Deep North
    Jul 17, 2015 at 1:52
  • $\begingroup$ NB: \lim_{y\to0^+} will give you $\lim_{y\to0^+}$, rather than what you have. $\endgroup$
    – Kyle Kanos
    Jul 17, 2015 at 1:53
  • $\begingroup$ How about other cases? (Say when the integral goes from zero to infinity) $\endgroup$
    – texmex
    Jul 17, 2015 at 1:54
  • $\begingroup$ @KyleKanos Indeed then is the above Integral valid say when we go from zero to infinity? $\endgroup$
    – texmex
    Jul 17, 2015 at 1:56
  • 3
    $\begingroup$ You should note when you've simultaneously asked the same question elsewhere: math.stackexchange.com/questions/1364032/…. $\endgroup$
    – JimB
    Jul 17, 2015 at 3:52

1 Answer 1


The equalities never hold.

The following analysis illustrates the benefits of simplifying complicated expressions before attempting to analyze them.

Because $\Phi$ is the CDF of a continuous variable supported everywhere, it is strictly increasing and nonzero. One way to write this fact is

$$\Phi(z + \lambda) / \Phi(z) \gt 1$$

for any all $z$ and any $\lambda \gt 0$. Applying this to

$$z = \frac{-\log(-k/y) + \mu_x}{\sigma_x}$$

(assuming $k \gt 0$ in the first integral, where $y\lt 0$, and $k \lt 0$ in the second integral where $y\gt 0$, for otherwise the logarithm is not defined) and

$$\lambda = \sigma_x \gt 0,$$

and observing that the remaining factor $y f(y)$ has a constant sign in either integral, allows us to replace everything in braces $\{\cdots \}$--which is precisely $\Phi(z+\lambda)/\Phi(z)$--by the strictly smaller value $1$, achieving a strict inequality.

Since in (1) $yf(y) \lt 0$ and in (2) $yf(y) \gt 0$, the integrand on the left hand side of (1) is strictly greater than the integrand on the right hand side and the integrand of the lhs of (2) is strictly less than the integrand of its rhs. Integration over any set of positive measure (which is the case here) will preserve such a strict inequality.

Therefore, there are no possible values of $k$, $\mu_x$, or $\sigma_x$ that make either (1) or (2) true.

  • $\begingroup$ Appreciative of your clarifications as always. Any suggestions on how we can simplify the two identities above? Also, the part where the logarithm does not hold, how would we tackle that. Please note, the above problems comes up in a related question. Also, in the original question, I am not sure if the simplification regarding the conditional density is correct. Could you please eyeball the steps at the answer here: stats.stackexchange.com/questions/157954/… $\endgroup$
    – texmex
    Jul 18, 2015 at 10:59

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