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It sounds like you want the formula for the density transformation under a monotonic function. If you start with a continuous random variable $X$ and you define the random variable $Y=f(X)$ using a s …

Short answer: Back-transformed coeffient estimator is biased, and not a good estimator. Back-transformed confidence interval is valid, but sub-optimal. Longer answer: Since you have not included a …

In this context $\log$ means $\ln$ (i.e., a logarithm with base $e$), so the second website is probably not claiming that it is a logarithm with a base of ten. For a logarithm with base ten you would …

Your basic reasoning is correct, but your expression for the variance of the process is wrong. If we let $Z_t \equiv \log Y_t$ then the process $\{ Z_t | t \in \mathbb{Z} \}$ is a standard Gaussian $ …

If you have $n=100000$ sample values then the two methods should give you estimated values that are very close to one another. Since you haven't supplied your calculations I cannot say what went wron …

It is not really anything to do with trying to get a normal distribution (and in most models, you don't really need to transform variables to make them normal anyway). We generally apply a logarithmi …

This largely depends on the robustness of your inferences to errors in the distribution When you are dealing with quantities that are either directly observable, or for which there is some close estim …

The numerator in your expression for $R_t$ is the logarithm of a weighted sum of lognormal random variables. The distribution of this quantity is complicated, but it has some known approximations (se …

Since the returns involve changes in the stock price over consecutive time periods, the answer to your question depends on the joint distribution of the stock price over time. Since you have only spe …

It appears that you are confusing the random variables $X_1,X_2,...,X_N$ with their expected values. As presently defined, the value $S$ is not a random variable at all; it is a constant (and so the …

The hyperbolic tangent function $\tanh$ is a strictly increasing function, so it is quite simple to get the CDF of the random variable $Y$. I am going to give a more general answer that what you are …

The intuition for this result comes from the fact that the exponential function is a strictly convex function. When you then impose a convex transformation on the random variable $X$, the positive de …