# Expectations of multiplied Wiener processes.

I wish to evaluate the following:

• $E[W(t-1)W(t)^2]$

• $E[W(t)^3]$

where $t > 1$, $W$ is a standard Brownian motion and we are at $\mathscr{F}_0$ now.

I know that $E[W(t-1)W(t)] = \min{(t-1,t)} = (t-1)$ when $W$ is a standard Brownian motion, but I'm not sure how to solve the above expectations. I could do it easily if I could go $E[W(t-1)W(t)^2]=E[W(t-1)]E[W(t)^2]=0$ (duh); however I can't do this beacuse they're dependent random variables.

Thanks.

-

## 1 Answer

For the first one, let $X = W(t) - W(t-1)$ and write $W(t-1)W(t)^2 = W(t-1)(X + W(t-1))^2$. After expanding the square you have three terms, each of which is a product of independent random variables.

For the second one: all you need to know is that $W(t)$ is normally distributed with mean 0 and variance $t$, so you know its density function, and can compute the expectation directly from that (integration by parts will help).

-

1
I beg to differ for the second one: the distribution of `W(t)` is symmetric hence `E[u(W(t))]=0` for every odd function `u`, for example `u(x)=x^3`. – Did Dec 13 '12 at 6:34
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9403489232063293, "perplexity_flag": "head"}