Expected Value of Continuous Random Distribution

Outcomes of experiments or values of a random variable are observed in order to discover whether there is a value or range of values that tends to occur more often than other values.

Sometimes the observations are clustered tightly around a central value while on other occasions, the values are more evenly spread across their range. In either case, there is a sense of a location for the values of the random variable.

When dealing with data generated by trials of an experiment or in an observational study, we take the mean value of the data as the best estimate of the location of the centre of the underlying distribution. This central value is called, in the language of statistics, the expected value of the random variable.

For a random variable $X$X, we notate its expected value by $E(X)$E(X). 

If we are not dealing directly with data but instead with an assumed probability distribution for a random variable, the expected value can be understood as an average value weighted by the probabilities for the possible values.

 

Discrete distribution

In the case of a discrete random variable $Y$Y, this means, for example, that if $Y=1,3,5,7$Y=1,3,5,7 with probabilities $\frac{1}{8},\frac{1}{4},\frac{3}{8}$18​,14​,38​ and $\frac{1}{4}$14​ respectively, then

$E(Y)$E(Y)	$=$=	$\frac{1}{8}\times1+\frac{1}{4}\times3+\frac{3}{8}\times5+\frac{1}{4}\times7$18​×1+14​×3+38​×5+14​×7
	$=$=	$4\frac{1}{2}$412​

One way to see this is to imagine a sample of, say, $1000$1000 observations of the random variable $Y$Y. According to the probabilities given, we would expect the value $1$1 to occur $125$125 times, $3$3 to occur $250$250 times, $5$5 to occur $375$375 times and $7$7 to occur $250$250 times. Averaging all these outcomes gives $\frac{125\times1+250\times3+375\times5+250\times7}{1000}$125×1+250×3+375×5+250×71000​. But this is exactly the same calculation as was done above.

 

Continuous distribution

In the case of a continuous random variable, the expected value is defined analogously in terms of an integral. If the random variable $X$X has probability density function $f$f that is defined on a domain $(a,b)$(a,b), we define

$E(X)=\int_a^bxf(x)\mathrm{d}x$E(X)=∫ba​xf(x)dx

 

Example 1

A certain event can occur at any time within $10$10 minutes of the starting time of an experiment and the random variable $T$T representing the time at which the event occurs is assumed to have a uniform probability distribution. 

Intuition suggests that the values of the random variable will be centred around the $5$5-minute mark. We can confirm this using the definition of expected value for a continuous random variable. 

The random variable $T$T has PDF  $f(T)=\frac{1}{10}$f(T)=110​. Therefore, $E(T)=\int_0^{10}\ t\frac{1}{10}\mathrm{d}t$E(T)=∫100​ t110​dt. This is, $\left[\frac{t^2}{20}\right]_0^{10}=5$[t220​]100​=5, as expected.

 

Example 2

A random variable $X$X has PDF given by $f(X)=\frac{1}{5}-\frac{X}{50}$f(X)=15​−X50​ where $X$X is in the interval $(0,10)$(0,10). What is the mean or expected value of $X$X?

By the definition $E(X)=\int_0^{10}\ x\left(\frac{1}{5}-\frac{x}{50}\right)\mathrm{d}x$E(X)=∫100​ x(15​−x50​)dx. So, 

$E(X)$E(X)	$=$=	$\int_0^{10}\ \left(\frac{x}{5}-\frac{x^2}{50}\right)\mathrm{d}x$∫100​ (x5​−x250​)dx
	$=$=	$\left[\frac{x^2}{10}-\frac{x^3}{150}\right]_0^{10}$[x210​−x3150​]100​
	$=$=	$10-\frac{1000}{150}$10−1000150​
	$=$=	$3\frac{1}{3}$313​

Worked Examples

Question 1

Question 2

Question 3