What is the expectation of W multiplied by the exponential of W?

Compute $\mathbb{E} [W_t \exp W_t] $.

Quantitative Finance Interviews are comprised of finance, programming and probability questions, as well as, some logic questions, known as brainteasers. In this post series, I share some frequently asked questions from Quantitative Finance Interviews for quantitative analysts with junior level of experience.

Here, I present a question on probability. The purpose with this question is to assess your knowledge on the Brownian motion (possibly on the Girsanov theorem).

Interview Question

Here is the question about the expectation of a function of the Brownian motion:

Let $(W_t)_{t>0}$ be a Brownian motion. Compute $\mathbb{E} [ W_t \exp W_t ]$.

Question Explanation

First, you need to understand what is a Brownian motion $(W_t)_{t>0}$. In fact, a Brownian motion is a time-continuous stochastic process characterized as follows:

  1. $W_0 = 0$;
  2. $W_t$ is continuous with probability 1;
  3. $W_{t_2} - W_{s_2}$ and $W_{t_1} - W_{s_1}$ are independent random variables for $0 \le s_1 < t_1 \le s_2 < t_2 $;
  4. $W_t - W_s \sim \mathcal{N}(0, t-s)$ for $0 \le s \le t$.

So, you need to use appropriately the Property 4, i.e., $W_t \sim \mathcal{N}(0,t)$. Since you want to compute the expectation of two terms where one of them is the exponential of a Brownian motion, it would be interesting to know $\mathbb{E} [\exp X]$, where $X$ is a normal distribution.

Moment-generating function

We define the moment-generating function $M_X$ of a real-valued random variable $X$ as $$ M_X (u) := \mathbb{E} [\exp (u X) ], \quad \forall u \in \mathbb{R}. $$

Let $Z$ be a standard normal distribution, i.e. $Z \sim \mathcal{N}(0,1)$. For some reals $\mu$ and $\sigma>0$, we build $X$ such that $X =\mu + \sigma Z$, i.e. $X \sim \mathcal{N}(\mu,\sigma^2)$. The moment-generating function $M_X$ is given by $$ M_X (u) = \mathbb{E} [\exp (u X) ] = \exp \big( \mu u + \tfrac{1}{2}\sigma^2 u^2 \big). $$

Answer I

By using the moment-generating function expression for $W\sim\mathcal{N}(0,t)$, we get: $$ M_{W_t} (u) = \mathbb{E} [\exp (u W_t) ] = \exp \big( \tfrac{1}{2} t u^2 \big). $$

Then, by differentiating the function $M_{W_t} (u)$ with respect to $u$, we get: $$ \tfrac{d}{du} M_{W_t}(u) = \tfrac{d}{du} \mathbb{E} [\exp (u W_t) ] = \mathbb{E} \big[ \tfrac{d}{du} \exp (u W_t) \big]
= \mathbb{E} \big[ W_t \exp (u W_t) \big] $$ and $$ \tfrac{d}{du} M_{W_t}(u) = \tfrac{d}{du} \exp \big( \tfrac{1}{2} t u^2 \big) = \tfrac{1}{2} t \exp \big( \tfrac{1}{2} t u^2 \big) \tfrac{d}{du} u^2 = t u \exp \big( \tfrac{1}{2} t u^2 \big) $$

From both expressions above, we have: $$ \mathbb{E} \big[ W_t \exp (u W_t) \big] = t u \exp \big( \tfrac{1}{2} t u^2 \big). $$ Taking $u=1$ leads to the expected result: $$ \mathbb{E} \big[ W_t \exp W_t \big] = t \exp \big( \tfrac{1}{2} t \big). $$


Using the idea of the solution presented above, the interview question could be extended to:

Let $(W_t)_{t>0}$ be a Brownian motion. Compute $\mathbb{E}[W_t^n \exp W_t]$ for every $n \ge 1$.


Isaque Pimentel
Quantitative Analyst, Consultant

Ph.D. in Applied Mathematics interested in Quantitative Finance and Data Science.