Prediction Interval for $E(y|X_0)$ when model is in $ln(y)$ Let's suppose we have linear regression with dependent variable $ln(y)$. How would one find a prediction interval for  $E(y|X_0)$ ?

If we have $PI(E(ln(y)|X_0))$, is there a way to reach $PI(E(y|X_0))$?

Any help would be appreciated.
I got hold of this link from Journal of Statistics Education

Using the notation in it, let $X\sim \log N(\mu,\sigma^2)$, i.e. $Y=\log(X) \sim N(\mu,\sigma^2)$.

Hence, $E(X)=E(e^Y)=e^{\mu+\sigma^2/2}$. An estimator for $\log(E(X))$ is $$\bar Y+\frac{S_Y^2}{2}$$. Also, $Var(\hat\log(E(X)))=\frac{\sigma^2}{n}+\frac{\sigma^4}{2(n-1)}$, which gives $$\hat Var(\hat\log(E(X)))=\frac{S_Y^2}{n}+\frac{S_Y^4}{2(n-1)}$$.

For large samples, using the CLT, we have $$\frac{\hat \log E(X)-\log E(X)}{\sqrt{\hat Var(\hat\log(E(X)))}}\sim^a N(0,1)$$

Thus, one possible CI from this perspective is $$CI(E(X))=\left]e^{\bar Y+\frac{S_Y^2}{2}\pm z_{1-\alpha}\sqrt{\frac{S_Y^2}{n}+\frac{S_Y^4}{2(n-1)}}}\right[$$.
According to the link, this is the Cox method, while the modified Cox method substitutes the standard normal quantile for a $t(n)$ quantile. They simulate more than just the these two intervals. They also simulate the naive approach.

If we look at table 5  of the link, we can see that the best method is the modified Cox, since for small samples, and large ones, the covering probability is the most close to the nominal level.

Alecos answer, if I understood correctly, suggests adding a correction term to the naive approach, resulting in $CI(E(X))=\left]e^{\bar Y+\frac{1}{2}\frac{S_X^2}{(\bar X)^2}\pm t_{1-\alpha}\sqrt{\frac{S_Y^2}{n}}}\right[$.

I've simulated his answer with the following code in Mathematica:

sim[n_] := Module[{},
  ct = 1;
  res = 0;
  While[ct &lt;= 10000,
   data = RandomVariate[LogNormalDistribution[5, 1], n];
   y = Log[data];
   my = Mean[y];
   s2y = Variance[y];
   mx = Mean[data];
   s2x = Variance[data];
   qt = Quantile[StudentTDistribution[n], 0.975];
   If[E^(my + 0.5*s2x/mx^2 - qt*Sqrt[s2y/n]) &lt; E^(5 + 0.5) &lt; 
     E^(my + 0.5*s2x/mx^2 + qt*Sqrt[s2y/n]),
    res++];
   ct++;
   ];
  res = res/10000.
  ]


The only difference is that I've done 10000 replications whereas the link only does 1000. 

Running the code for the same sample sizes of the link, we obtain the following covering probabilities: $${0.8954, 0.8464, 0.7682, 0.7153, 0.6454, 0.4832, 0.3045, 0.073, 0.0068}$$.

AS we can see the correction increases the covering probabilities when compared to the naive method, but it still suffers from the same problem, i.e., the prob. tend to zero as the sample size increases.

All in all, the best course of action is to use the modified Cox approach.