# 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:

- $W_0 = 0$;
- $W_t$ is continuous with probability 1;
- $W_{t_2} - W_{s_2}$ and $W_{t_1} - W_{s_1}$ are independent random variables for $0 \le s_1 < t_1 \le s_2 < t_2 $;
- $W_t - W_s \sim \mathcal{N}(0, t-s)$ 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: $$ \mathbb{E} \big[ W_t \exp W_t \big] = t \exp \big( \tfrac{1}{2} t \big). $$

## Applications

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 $n \ge 1$.