Expectation : Exercises

Introduction

In the following problems the aim is (almost) always to find the expectation for a random variable. Remember that for discrete random variable the expectation is given by: $$ \mathbb{E}(X) = \sum_{x_i=0}^\infty x_i P(X=x_i) = \sum_{x_i=0}^\infty x_i f_X(x_i) $$ where $f_X(x)$ is used to denote the probability mass function. For a continuous random variable the expectation is: $$ \mathbb{E}(X) = \int_{-\infty}^\infty x f_X(x) \textrm{d}x $$ where $f_X(x)$ is the probability density. If the expectation of a function of a random variable $\mathbb{E}[g(X)]$ is required then this can be calculated using: $$ \mathbb{E}[g(X)] = \sum_{x_i=0}^\infty g(x_i) f_X(x_i) $$ in the discrete case and using: $$ \mathbb{E}[g(X)] = \int_{-\infty}^\infty g(x) f_X(x) \textrm{d}x $$ in the continuous case.

Example problems

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Problem 1

The discrete random variable, $X$, has the values: $X \in \{ 0,1,2,3\}$, with a probability mass given by: $$ f_X(x) = P(X=x) =k x (x-1) \quad , \quad x=0,1,2,3 \quad . $$ where $k$ is a constant. Find the value of $k$ and then find the expectation, $\mathbb{E}(X)$, and variance, $\textrm{var}(X)$, of this random variable.

To work out $k$ we can use the fact that the distribution must be normalised. In other words, we use the fact that: $$ \sum_{i=0}^\infty f_X(x) = 1 $$ In this case this implies: $$ \sum_{i=0}^3 f_X(x) = 1 $$ Using the definition of $f_X(x)$ in the question this implies: $$ 0 + 0 + 2 k + 6 k = 1 \qquad \rightarrow \qquad k =\frac{1}{8} $$ The expectation of a discrete random variable is given by: $$ \mathbb{E}(X) = \sum_{i=0}^\infty x f_X(x) $$ In this case we have four terms in this sum: $$ \mathbb{E}(X) = 0 \times 0 + 1 \times 0 + 2 \times \frac{1}{4} + 3 \times \frac{3}{4} = \frac{11}{4} $$ The variance is given by: $$ \textrm{var}(X) = \mathbb{E}(X^2) - [\mathbb{E}(X)]^2 \qquad \textrm{where} \qquad \mathbb{E}(X^2) = \sum_{i=0}^\infty x^2 f_X(x) $$ In this case $\mathbb{E}(X^2)$ is given by : $$ \mathbb{E}(X^2) = 4 \times \frac{1}{4} + 9 \times \frac{3}{4} = \frac{31}{4} $$ The variance is thus: $$ \textrm{var}(X) = \frac{31}{4} - \left( \frac{11}{4} \right)^2 = \frac{3}{16} $$

Problem 2

Suppose that the value/price of an asset (for example a share in a company) changes randomly over time. In the geometric Brownian motion model, its value, $S$, at a time, $t$, is given by $$ S(t)=S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t +\sigma X \sqrt{t} \right) \qquad , $$ where $\mu$ is the drift rate, $\sigma$, the volatility, and where $X$ is a random variable, with the standard distribution, $N(0,1)$. Clearly, $S_{0}$, is the initial ($t=0$) value of the asset. Show that the expected value of the asset, after a time $t$, is given by: $$ \mathbb{E}[S(t)] = \frac{ S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t \right) }{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left( \sigma x \sqrt{t} - \frac{x^2}{2} \right) \textrm{d}x $$ Then use the fact that $\sigma x \sqrt{t} - \frac{x^2}{2} = -\frac{1}{2}(x-\sigma\sqrt{t})^2 + \frac{1}{2}\sigma^2 t$, the substitution $y = x - \sigma \sqrt{t}$ and the fact that $\int_{-\infty}^\infty e^{-y^2/2} \textrm{d}y = \sqrt{2\pi}$ to show that the value of the above integral is nothing more than: $$ \mathbb{E}[S(t)] = S_0 \exp(\mu t) $$

The definition of the expectation for a continuous random variable : $\mathbb{E}[S(t)] = \int_{-\infty}^\infty s(t) P\left\{ S(t) = s \right\} \textrm{d}s(t)$

Now from the question: $$ P\{ S(t)=s \} = S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t +\sigma X \sqrt{t} \right) = S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t \right)\exp\left( \sigma X \sqrt{t} \right) $$ We are calculating $\mathbb{E}[S(t)]$ at fixed $t$ hence the first exponential in the above is fixed. This is important as we only need to keep the terms that depend on the random variable, $X$, inside the integral for the expectation. When we take this integral we are going to integrate over the set of values that this random variable can take. Now the question tells us that $$ P\{X = x \} = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \qquad \qquad \textrm{Normal distribution} $$ Given all of the above we can rewrite our expectation value as: $$ \mathbb{E}[S(t)] = \int_{-\infty}^{\infty} s(t,x) P\{ X=x \} \textrm{d}x = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} s(t,x) e^{-x^2/2} \textrm{d}x $$ We can now insert $s(t,x)$ into the integral and manipulate this expression somewhat $$ \begin{aligned} \mathbb{E}[S(t)] & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t +\sigma x \sqrt{t} \right) e^{-x^2/2} \textrm{d}x \\ & = \frac{ S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t \right) }{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left( \sigma x \sqrt{t} - \frac{x^2}{2} \right) \textrm{d}x \end{aligned} $$ For the next step of the solution we have to note that : $\sigma x \sqrt{t} - \frac{x^2}{2} = - \frac{1}{2}(x - \sigma \sqrt{t})^2 + \frac{1}{2} \sigma^2 t$. We can get this result by competing the square, which is an algebraic trick that it is useful to know about. In this question we were given this information in the question. $$ \begin{aligned} \mathbb{E}[S(t)] & = \frac{ S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t \right) }{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left( \frac{1}{2}(x - \sigma \sqrt{t})^2 + \frac{1}{2} \sigma^2 t \right) \textrm{d}x \\ & = \frac{ S_0 \exp \left( t(\mu-\frac{1}{2}\sigma^2) + \frac{1}{2} \sigma^2 t \right)}{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left( \frac{1}{2}(x - \sigma \sqrt{t})^2 \right) \textrm{d}x \\ & = \frac{S_0 \exp\left( \mu t \right) }{\sqrt{2\pi}} \int_{-\infty}^\infty \exp\left( \frac{1}{2}(x - \sigma \sqrt{t})^2 \right) \textrm{d}x \end{aligned} $$ Now we finish the integration by making the substitution $y = x - \sigma \sqrt{t} \quad \rightarrow \quad \frac{\textrm{d}y}{\textrm{d}x } = 1$ Remember when we do an integration by substitution we have to change the limits, the integrand and the infinitessimal (the $\textrm{d}x$ part). $$ \mathbb{E}[S(t)] = \frac{S_0 \exp\left( \mu t \right) }{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-y^2/2} \textrm{d}y = \frac{S_0 \exp\left( \mu t \right) }{\sqrt{2\pi}} \sqrt{2\pi} = S_0 \exp\left( \mu t \right) $$ It is worth noting something about this problem and its final solution. This problem is typical of the sort of thing that appears in financial maths. As discussed in the question this particular random variable appears in the context of modelling the value/price of an asset. Now the value of an asset changes for two reasons (1) the value increases geometrically because of compound interest and (2) the value changes randomly because of the various random factors that affect the assets price. The model has therefore been constructed so that on average it increases geometrically. There might, however, be some random fluctuations around this average. As such many of the terms that appear in the original random variable: $$ S(t)=S_0 \exp \left( (\mu-\frac{1}{2}\sigma^2)t +\sigma X \sqrt{t} \right) $$ (particularly the $-\frac{1}{2}\sigma^2 t$ that appears in the exponential) are there so that the expectation comes out as: $$ \mathbb{E}[S(t)] = S_0 \exp\left( \mu t \right) $$ I mention all this stuff because it is interesting and because we often don't get students to think about how we might (when modelling) set up ``problems" so that the solution comes out in a particular way.

Expectation : Exercises

Introduction

Example problems

Problem 1

Problem 2

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