Difference of two i.i.d. lognormal random variables Let $X_1$ and $X_2$ be 2 i.i.d. r.v.'s where $\log(X_1),\log(X_2) \sim N(\mu,\sigma)$. I'd like to know the distribution for $X_1 - X_2$.
The best I can do is to take the Taylor series of both and get that the difference is the sum of the difference between two normal r.v's and two chi-squared r.v.'s in addition to the rest of the difference between the rest of the terms. Is there a more straight-forward way to get the distribution of the difference between 2 i.i.d. log-normal r.v.'s?
 A: This is a difficult problem.  I thought first about using (some approximation of) the moment generating function of the lognormal distribution.  That doesn't work, as I will explain. But first some notation:
Let $\phi$ be the standard normal density and $\Phi$ the corresponding cumulative distribution function.  We will only analyze the case lognormal distribution $lnN(0,1)$, which has density function
$$
     f(x)=\frac1{\sqrt{2\pi}x} e^{-\frac12 (\ln x)^2} 
$$
and cumulative distribution function
$$
      F(x) =\Phi(\ln x)
$$
Suppose $X$ and $Y$ are independent random variables with the above lognormal distribution.  We are interested in the distribution of $D=X-Y$, which is a symmetric distribution with mean zero.  Let $M(t) = \DeclareMathOperator{\E}{E}  \E e^{tX} $ be the moment generating function of $X$.  It is defined only for $t\in (-\infty,0]$, so not defined in an open interval containing zero.  The moment generating function for $D$ is $M_D(t)=\E e^{t(X-Y)}= \E e^{tX} \E e^{-tY}= M(t)M(-t)$.  So, the moment generating function for $D$ is only defined for $t=0$, so not very useful. 
That means we will need some more direct approach for finding approximations for the distribution of $D$.  Suppose $t\ge 0$, calculate
$$
   \begin{align} 
       P(D \le t) &= P(X-Y\le t) \\
                  &= \int_0^\infty P(X-y\le t | Y=y) f(y) \; dy \\
                  &= \int_0^\infty P(X\le t+y) f(y) \; dy  \\
                  &= \int_0^\infty F(t+y) f(y) \; dy
   \end{align}
$$ (and the case $t<0$ is solved by symmetry, we get $P(D\le t)=1-P(D\le |t|)$).      
This expression can be used for numerical integration or as a basis for simulation.  First a test:
 integrate(function(y) plnorm(y)*dlnorm(y), lower=0,  upper=+Inf)
  0.5 with absolute error < 2.3e-06

which is clearly correct.  Let us wrap this up inside a function:
pDIFF  <-  function(t) {
    d  <-  t
    for (tt in seq(along=t)) {
        if (t[tt] >= 0.0) d[tt] <- integrate(function(y) plnorm(y+t[tt])*dlnorm(y),
                                         lower=0.0,  upper=+Inf)$value else
                          d[tt] <- 1-integrate(function(y) plnorm(y+abs(t[tt]))*dlnorm(y),
                                         lower=0.0, upper=+Inf)$value
    }
    return(d)
}

> plot(pDIFF,  from=-5,  to=5)

which gives:

Then we can find the density function by differentiating under the integral sign, obtaining
dDIFF  <-  function(t) {
       d  <- t; t<- abs(t)
       for (tt in seq(along=t)) {
           d[tt]  <-  integrate(function(y) dlnorm(y+t[tt])*dlnorm(y),
                                lower=0.0,  upper=+Inf)$value
       }
       return(d)
}

which we can test:
> integrate(dDIFF,  lower=-Inf,  upper=+Inf)
0.9999999 with absolute error < 1.3e-05

And plotting the density we get:
plot(dDIFF,  from=-5,  to=5)


I did also try to get some analytic approximation, but so far didn't succeed, it is not an easy problem.  But numerical integration as above, programmed in R is very fast on modern hardware, so is a good alternative which probably should be used much more.
A: This does not strictly answer your question, but wouldn't it be easier to look at the ratio of the $X$ and $Y$? You then simply arrive at
$$
\begin{align}
\Pr\left(\frac{X}{Y} \leq t\right) &= \Pr\left(\log\left(\frac{X}{Y}\right) \leq \log(t) \right) \\
&= \Pr(\log(X) - \log(Y) \leq \log(t)) \\
&\sim \mathcal{N}(0, 2 \sigma^2)
\end{align}$$
Depending on your application, this may serve your needs.
A: I have a feeling characteristic functions are the best option for handling this question.
For basics, kindly see:https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)
now, kindly CTRL+F "independen". you'll see the answer
in brief:

*

*The characteristic function approach is particularly useful in analysis of linear combinations of independent random variables


*
If X1, ..., Xn are independent random variables, and a1, ..., an are some constants, then the characteristic function of the linear combination of the Xi 's is

  φ a 1 X 1 + ⋯ + a n X n ( t ) = φ X 1 ( a 1 t ) ⋯ φ X n ( a n t ) . {\displaystyle \varphi _{a_{1}X_{1}+\cdots +a_{n}X_{n}}(t)=\varphi _{X_{1}}(a_{1}t)\cdots \varphi _{X_{n}}(a_{n}t).} 



*
One specific case is the sum of two independent random variables X1 and X2 in which case one has

φ X 1 + X 2 ( t ) = φ X 1 ( t ) ⋅ φ X 2 ( t ) . {\displaystyle \varphi {X{1}+X_{2}}(t)=\varphi {X{1}}(t)\cdot \varphi {X{2}}(t).}


*Characteristic functions are particularly useful for dealing with linear functions of independent random variables. For example, if X1, X2, ..., Xn is a sequence of independent (and not necessarily identically distributed) random variables, and
S n = ∑ i = 1 n a i X i , {\displaystyle S_{n}=\sum {i=1}^{n}a{i}X_{i},,!} S_{n}=\sum {i=1}^{n}a{i}X_{i},,!
where the ai are constants, then the characteristic function for Sn is given by
φ S n ( t ) = φ X 1 ( a 1 t ) φ X 2 ( a 2 t ) ⋯ φ X n ( a n t ) {\displaystyle \varphi _{S_{n}}(t)=\varphi _{X_{1}}(a_{1}t)\varphi _{X_{2}}(a_{2}t)\cdots \varphi _{X_{n}}(a_{n}t)\,\!} \varphi _{S_{n}}(t)=\varphi _{X_{1}}(a_{1}t)\varphi _{X_{2}}(a_{2}t)\cdots \varphi _{X_{n}}(a_{n}t)\,\!

In particular, φX+Y(t) = φX(t)φY(t). To see this, write out the definition of characteristic function:
φ X + Y ( t ) = E ⁡ [ e i t ( X + Y ) ] = E ⁡ [ e i t X e i t Y ] = E ⁡ [ e i t X ] E ⁡ [ e i t Y ] = φ X ( t ) φ Y ( t ) {\displaystyle \varphi _{X+Y}(t)=\operatorname {E} \left[e^{it(X+Y)}\right]=\operatorname {E} \left[e^{itX}e^{itY}\right]=\operatorname {E} \left[e^{itX}\right]\operatorname {E} \left[e^{itY}\right]=\varphi _{X}(t)\varphi _{Y}(t)} {\displaystyle \varphi _{X+Y}(t)=\operatorname {E} \left[e^{it(X+Y)}\right]=\operatorname {E} \left[e^{itX}e^{itY}\right]=\operatorname {E} \left[e^{itX}\right]\operatorname {E} \left[e^{itY}\right]=\varphi _{X}(t)\varphi _{Y}(t)}

The independence of X and Y is required to establish the equality of the third and fourth expressions.
