I don't get this, I am asked if the following $X_t$ is a brownian motion or not.
$Z$ is a standard normal variate. $X_t=\sqrt{t}Z$. I s$X_t$ a Brownian motion?
Answer is apparently no and one of the reasons is that the variance of $X_t-X_s=$ is $(\sqrt{t}-\sqrt{s})^2$. But I get $t-s$ as the variance of the increment. Here's how
$var(X_t-X_s)=\mathbb{E}[(X_t-X_s)^2]-\mathbb{E}[X_t-X_s]^2$ say $0<s<t$.
Then, $\mathbb{E}[X_t^2-2X_tX_s+X_s^2]-(\mathbb{E}[X_t]-\mathbb{E}[X_2])^2 = \mathbb{E}[x_t^2]-2\mathbb{E}[X_tX_s]+\mathbb{E}[x_s^2]-(0)^2$ since $\mathbb{E}[X_t]=\sqrt{t}\mathbb{E}[Z]$ and $Z$ is a standard normal variable. Same for $X_s$.
So $\mathbb{E}[X_t^2]=t \mathbb{E}[Z^2]$. $Z$ is standard normal so $var(Z)=\mathbb{E}[Z^2]=1$. I calculate $\mathbb{E}[X_tX_s]$ using that $X_s-X_0$ and $X_t-X_s$ are independent. Then $\mathbb{E}[X_s]\mathbb{E}[X_t-X_s]+\mathbb{E}[X_s^2]=s$ is what I get and substituting back, my overall variance is $t-2s+s=t-s$
So yes, the variance is $t-s$ and not $(\sqrt{t}-\sqrt{s})^2$. What did I do wrong? I cannot find a fault in my argument. Please tell me.
Thank you
Consider the sample path or realization of the process $\{X_t\colon t \geq 0\}$ corresponding to the outcome $Z = z$. It is the curve $\sqrt{t}z\mathbf 1_{[0,\infty)}(t)$ passing through the points $(0,0)$ and $(1,z)$. Does this resemble Brownian motion at all?
More specifically and in response to your question as to where you went wrong in your calculations, let us look at your claim that
Since $X_0 = 0$, we have that $X_s-X_0 = X_s = \sqrt{s}Z$ while $X_t-X_s = \sqrt{t}Z-\sqrt{s}Z = \left(\sqrt{t}-\sqrt{s}\right)Z$ both are multiples of $Z$ and so cannot be independent. Indeed, since $Z$ is a standard normal random variable, it is straightforward to get that the variance of $X_t-X_s = \left(\sqrt{t}-\sqrt{s}\right)Z$ is $\left(\sqrt{t}-\sqrt{s}\right)^2$ as your book claims, and not $(t-s)$ as you calculated.