let $W_t$ be a standard Brownian motion,
how can I calculate $E(W_t)$, I know the standard Brownian motion is symmetric, so this would be $0$ but how could I prove it via normal distribution?
and also $E(e^{t+W_t})$.
Calculating expectation of standard Brownian motion $W_t$
brownian motiondensity functionexpected value
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Let's recall that a Brownian bridge $(U_t)_{t \in [0,1]}$ is a Gaussian process with continuous sample paths, mean zero and covariance function
$$C(s,t) := \mathbb{E}(U_s \cdot U_t) = s \wedge t - s \cdot t \tag{1}$$
In order to show that $W_t := U_t+t \cdot Z$ is a Brownian motion we have to check the following properties:
- $W_0 = 0$ a.s.
- $t \mapsto W_t(\omega)$ is continuous for almost all $\omega$
- $(W_t)_{t \in [0,1]}$ is a Gaussian process such that the vector $\Gamma := (W_{t_1},\ldots,W_{t_n})$ is Gaussian with mean zero and covariance matrix $(\text{cov}(W_{t_j},W_{t_k}))_{j,k}=(t_j \wedge t_k)_{j,k}$ for any $0<t_1<\ldots<t_n<1$.
Proof:
- By definition, we have $U_0=0$ a.s. Obviously, this implies $W_0 = U_0+0=0$ a.s.
- Since $t \mapsto U_t(\omega)$ is continuous almost surely, we find that $t \mapsto W_t(\omega)$ is continuous almost surely.
- Let $0<t_1<\ldots<t_n<1$. Since $(U_t)_{t \in [0,1]}$ is a Gaussian process with mean $0$ and covarianc function $C$, we know that $\Lambda := (U_{t_1},\ldots,U_{t_n})$ is Gaussian with mean $0$ and covariance matrix $(C(t_j,t_k))_{j,k}$. Therefore, we find that $$\Gamma = \Lambda + \begin{pmatrix} t_1 \\ \vdots \\t _n \end{pmatrix} \cdot Z$$ is Gaussian as a sum of two independent Gaussian random variables. It remains to calculate the mean and covariance matrix. We have $$\mathbb{E}\Gamma = \mathbb{E}\Lambda + \begin{pmatrix} t_1 \\ \vdots \\t _n \end{pmatrix}\cdot \mathbb{E}Z = 0$$ since $\mathbb{E}\Lambda = 0$, $\mathbb{E}Z = 0$. Similarly, $$\begin{align*} \mathbb{E}(\Gamma_j \cdot \Gamma_k) &= \mathbb{E}(W_{t_j} \cdot W_{t_k}) = \mathbb{E}(U_{t_j} \cdot U_{t_k} + t_j \cdot Z \cdot U_{t_k} + t_k \cdot Z \cdot U_{t_j} + t_j \cdot t_k \cdot Z^2) \\ &= C(t_j,t_k) + t_j \underbrace{\mathbb{E}Z}_{0} \cdot \mathbb{E}U_{t_k} + t_k \cdot \underbrace{\mathbb{E}Z}_{0} \cdot \mathbb{E}U_{t_j}+t_j \cdot t_k \cdot \mathbb{E}(Z^2) \end{align*}$$ where we used the independence of $Z$ and $(U_t)_{t \in [0,1]}$. From $\mathbb{E}(Z^2)=1$ and $(1)$, it follows easily that $\text{cov}(W_{t_j},W_{t_k}) = t_j \wedge t_k$.
This finishes the proof of part (a). The proof of the second part is similarly, I leave it to you.
For each fixed $t$, the random variable $X_t=W_{2t}-W_t$ is centered normal with variance $t$ hence indeed distributed like $W_t$.
But the process $(X_t)$ is not a Brownian motion. To see this, note that $X_2=W_4-W_2$ and $X_1=W_2-W_1$ are the increments of a Brownian motion on some disjoint time intervals hence they are independent while $W_2$ and $W_1$ are not (for example, $E(X_2X_1)=0$ and $E(W_2W_1)=1$).
Best Answer
The key here is to note that the Brownian motion at time $t$ is distributed normally, with mean zero and variance (not standard deviation) $t$. To show that $\mathbb{E}[W_t] = 0$, you don't even need the fact about the variance. Just note that its mean is always zero.
To calculate $\mathbb{E}[W_t]$, note that you can first take $e^t$ out of the expectation (it is not a random variable). It remains to calculate $\mathbb{E}[e^{W_t}] = M(1)$, where $M$ is the moment generating function (MGF) of the random variable $W_t$. (You may recall that the definition of the MGF $M(\alpha)$ for a random variable $X$ is $\mathbb{E}[e^{\alpha X}]$.) We can again use the fact that $W_t$ is normally distributed. The MGF of a normal random variable is well-known: $$M(\mathcal{N}(\mu, \sigma^2)) = e^{\mu \alpha + \alpha^2 \sigma^2 / 2}$$ By taking $\mu = 0$, $\sigma^2 = t$, and $\alpha = 1$, we get that $M(W_t) = e^{t/2}$. Hence, $\mathbb{E}[e^{t + W_t}] = e^{3t / 2}$.