Expected value for continuous probability distributions
Recall that the basic expected value formula is the probability of an event multiplied by the number of trials for a particular probability experiment. That is $P\left(X\right)\times x$P(X)×x where $X$X is the event and $x$xis the number of trials. The expected value of a random variable gives the mean of the distribution.
The notation used for the expected value of a random variable $X$X is $\mu=E\left(X\right)$μ=E(X)
For discrete probability distributions the formula for expected value is $E\left(X\right)=\Sigma\ xP\left(x\right)$E(X)=Σ xP(x)
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 $\left(a,b\right)$(a,b), we define
$E\left(X\right)=\int_a^b\ xf\left(x\right)\ \mathrm{d}x$E(X)=∫ba xf(x) dx
Worked examples
Example 1
A certain event can occur at any time $T$T within $10$10 minutes of the starting time of an experiment. The time at which the event occurs is assumed to have a uniform probability distribution.
Think: 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.
Do: As we have been told this is a uniform probability distribution, the random variable $T$T has PDF:
$f\left(T\right)=\frac{1}{10}$f(T)=110
Therefore:
Thus, the expected value is $5$5 as we predicted.
Example 2
A random variable $X$X has PDF given by $f\left(X\right)=\frac{1}{5}-\frac{X}{50}$f(X)=15−X50 where $X$X is in the interval $$. What is the mean or expected value of $X$X?
Solution: By the definition $E\left(X\right)=\int_0^{10}\ x\left(\frac{1}{5}-\frac{x}{50}\right)\mathrm{d}x$E(X)=∫100 x(15−x50)dx. So:
Calculating variance ($\sigma^2$σ2) and standard deviation ($\sigma$σ) in a continuous probability distribution
Recall that variance is technically the average of the squared differences of each data value from the mean. Variance gives us a general idea of how spread the data values in a distribution are.
Because the expected value, $E\left(X\right)$E(X), is equivalent to the mean of the random variable we can express the formula for variance in terms of $E\left(X\right)$E(X). For example for discrete probability distributions the variance is:
$Var\left(X\right)=E\left(\left(X-\mu\right)^2\right)=\Sigma\left(x-\mu\right)^2P\left(x\right)$Var(X)=E((X−μ)2)=Σ(x−μ)2P(x)
As you can expect for continuous probability distributions the formula for variance changes to:
$Var\left(X\right)=E((x-\mu)^2)=\int_a^b\ \left(x-\mu\right)^2\ f\left(x\right)dx$Var(X)=E((x−μ)2)=∫ba (x−μ)2 f(x)dx
For ease of calculating, we can rewrite this as: $Var\left(X\right)=E\left(X^2\right)-\mu^2$Var(X)=E(X2)−μ2 or $E\left(X^2\right)-E\left(X\right)^2$E(X2)−E(X)2
Remember also that standard deviation is the square root of the variance: $\sigma=\sqrt{Var}$σ=√Var
We can summarise the difference between discrete and continuous probability distributions in relation to variance and expected value below:
| Discrete Probability Distribution | Continuous Probability Distributions | |
|---|---|---|
| Expected Value | $E\left(X\right)=\Sigma xP\left(x\right)$E(X)=ΣxP(x) | $E\left(X\right)=\int_a^b\ xf\left(x\right)\ \mathrm{d}x$E(X)=∫ba xf(x) dx |
| Variance | $Var\left(X\right)=E\left(\left(X-\mu\right)^2\right)=\Sigma\left(x-\mu\right)^2P\left(x\right)$Var(X)=E((X−μ)2)=Σ(x−μ)2P(x) | $Var\left(X\right)=E\left(\left(x-\mu\right)^2\right)=\int_a^b\ \left(x-\mu\right)^2\ f\left(x\right)dx$Var(X)=E((x−μ)2)=∫ba (x−μ)2 f(x)dx |
Practice questions
Question 1
Consider the probability density function $p$p where $p\left(x\right)=\frac{1}{20}$p(x)=120 when $25\le x\le45$25≤x≤45 and $p\left(x\right)=0$p(x)=0 otherwise.
Use integration to determine the expected value of $p\left(x\right)$p(x).
Use integration to determine the variance of $p\left(x\right)$p(x).
Round your answer to two decimal places if necessary
Question 2
Consider the probability density function $p$p, where $p\left(x\right)=kx^2$p(x)=kx2 when $2\le x\le5$2≤x≤5 and $p\left(x\right)=0$p(x)=0 otherwise.
Use integration to determine the value of $k$k.
Round your answer to four decimal places if necessary.
Use integration to determine the expected value of a random variable $X$X if it is distributed according to $p\left(x\right)$p(x).
Round to four decimal places if necessary.
By performing an integration similar to the one in part (b), we can find that $E\left(X^2\right)=\frac{1031}{65}$E(X2)=103165.
Hence, calculate the variance, $V\left(X\right)$V(X), of $p\left(x\right)$p(x).
Round to four decimal places if necessary.