Invariance Properties of Brownian Motion

Question

I am trying to make sense of the Scaling-Invariance and Time-Inversion properties of Brownian motion by producing a sample path. For the record, I am using the following definitions. Let $B(t)$ be the standard Brownian Motion. One sample path is produced using random walk as shown as the black line. For Scaling-Invariance property, defind $X_t := \frac{1}{a}B(a^2t)$. This $X(t)$ is also supposed to be a standard Borwnian motion. This is shown as the blue line. For Time-Inversion, define $Y(t):=tB(1/t), t\neq 0$ and $Y(0) = 0$. This is shown as the red line.

By looking at the graph, the blue line makes sense. It basically is a fast track of the original Brownian motion with magnitude halved. However, the red line does not make sense to me. Did I produce the right picture, please? If so, how do I intepret the red line, please? Thank you!

> n <- 1000
> set.seed(n)
> t <- seq(0, 1, by=1/n)
> X <- c(0, rnorm(n))

> B1 <- cumsum(X) # BM from random walk
> B2 <- t*rev(B1) # Time-Inversion
> a <- 2
> t2 <- seq(0, 0.25, by=0.25/1000)
> B3 <- (1/a)*B1 # Scaling-Invariance
> plot(t, B1, type="l", ylab="BM")
> lines(t, B2, col=2)
> lines(t2, B3, col=4)
> legend("bottomleft", legend=c("BM", "BM-Scale-Invariance", "BM-Time-> Inversion"), lty=1, col=c(1, 4, 2))

Answer 1

The line B2 <- t*rev(B1) produces $t\mapsto t\cdot B(1-t)$, not $Y:t\mapsto t\cdot B(1/t)$. Two signs of what is going wrong are that B2 produces graphs that are (i) too flat near $t=0$ and (ii) quite similar near $t=1$ to a mirrored version of the B1 graphs near $t=0$.