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• Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T = B_T + T $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.
• And with the condition, it doesn't exist: with probability $1$, the Brownian motion will take negative values around $t=0$ (consequence of Proposition 9, p. 28).

Also, from the Arcsin Law:

Theorem : The probability that the Brownian motion $\left(B\right)_{t\ge0}$ has no zeros in the time interval $(a,b)$ is given by $\frac{2}{\pi}\arcsin\sqrt{\frac{a}{b}}$.

So for $a=0$, it is zero, and by independent increments the probability to remain positive in a neighborhood of $0$ is null.

• Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T \sim B_T + T, $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.
• And with the condition, it has probability $0$ to occur: the Brownian motion will take negative values around $t=0$ with probability $1$ (consequence of Proposition 9, p. 28).

Also, from the Arcsin Law:

Theorem : The probability that the Brownian motion $\left(B\right)_{t\ge0}$ has no zeros in the time interval $(a,b)$ is given by $\frac{2}{\pi}\arcsin\sqrt{\frac{a}{b}}$.

So for $a=0$, it is zero, and by independent increments the probability to remain positive in a neighborhood of $0$ is null.

Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T = B_T + T $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.

And with the condition, it doesn't exist: with probability $1$, the Brownian motion will take negative values around $t=0$ (consequence of Proposition 9, p. 28).

• Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T = B_T + T $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.
• And with the condition, it doesn't exist: with probability $1$, the Brownian motion will take negative values around $t=0$ (consequence of Proposition 9, p. 28).

Also, from the Arcsin Law:

Theorem : The probability that the Brownian motion $\left(B\right)_{t\ge0}$ has no zeros in the time interval $(a,b)$ is given by $\frac{2}{\pi}\arcsin\sqrt{\frac{a}{b}}$.

So for $a=0$, it is zero, and by independent increments the probability to remain positive in a neighborhood of $0$ is null.

Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T = B_T + T $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.

And with the condition, it doesn't exist: with probability $1$, the Brownian motion will take negative values around $t=0$.

Without the condition $0\le W_t\le1$ on $[0,1]$, it calls a Brownian bridge: $$ W'_T = B_T + T $$ where $B=\left(B_t\right)_{t\ge0}$ is the standard Brownian motion.

And with the condition, it doesn't exist: with probability $1$, the Brownian motion will take negative values around $t=0$ (consequence of Proposition 9, p. 28).