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?
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.
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.
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.
