Random variable $X$ is a **continuous random variable** if there is a function $f: \mathbb{R} \rightarrow [0, \infty>$ such that $$P(X \leq a) =\int_{-\infty}^{a} f(t) dt, \quad \forall a \in \mathbb{R}$$

Function $f$ is called the **probability density function** or **PDF**.

Continuous random variable can take on an uncountably infinite number of possible outcomes, therefor its name.

**Proposition.** Let $X$ be a continuous random variable with density $f$.

**(a)** $\forall a,b \in \mathbb{R}, a\leq b$ we have

$$P(a < X \leq b)=P( \{X\leq b\} \setminus \{X\leq a\})=P(x \leq b)-P(X \leq a) = \int_{a}^{b} f(t) dt$$

**(b)** All probabilities in measured space and up to $1$. In continuous variables integral is interpreted as sum.$$\int_{-\infty}^{+\infty} f(t)dt =P(X \in \mathbb{R}) =1$$

**(c)** Notice that for all $a \in \mathbb{R}$ $$P(X=a)=\int_{a}^{a} f(t)dt=0$$ Consequently, $$P(X \leq a)=P(X <a)+P(X=a)= P(X<a)$$ Moreover, $$P(a <X \leq b)=P(a <X<b)=P(a \leq X < b) =P(a \leq X \leq b)$$ i.e. it’s not important if the edge is included or not.

**Proposition.**

**(a)** **Expected value** of continuous random variable is defined as $$E[X]:= \int_{-\infty}^{+\infty} tf(t)dt$$

Generally, if $g:\mathbb{R} \rightarrow \mathbb{R}$ $$E[g(X)]:= \int_{-\infty}^{+\infty} g(t)f(t)dt$$

**(b) Variance** is the same as for discrete random variable, $$Var(X)=E[X^{2}]-(E[X])^{2}$$

#### Example 1

Let $X$ be continuous random variable with PDF:

$$f(t)=\begin{cases} c\cdot t^{4}, & \text{if $0<t<1$}.\\ 0, & \text{otherwise}.\end{cases}$$

**(a)** Find constant $c$

**(b)** Calculate $P(X>\frac{1}{2})$

**(c)** Find $E[X]$ and $Var(X)$.

**Solution**

**(a)** We know that $\int_{-\infty}^{+\infty} f(t)dt =1$. Since the function has value $0$ for every $t$ outside of the interval $(0,1)$ the integral is equal to $$\int_{0}^{1} c\cdot t^{4} dt =1$$

Further, lets solve this integral by using the polynomial rule

$$\displaystyle{\frac{c}{5}\cdot t^{5} \Big|_0^1=1 \Rightarrow c=5}$$

Consequently, PDF looks like this

$$f(t)=\begin{cases} 5\cdot t^{4}, & \text{if $0<t<1$}.\\ 0, & \text{otherwise}.\end{cases}$$

**(b)** Probability of $X>\frac{1}{2}$ can be calculated in two ways.

First way, $$P(X > \frac{1}{2})= \int_{-\infty}^{+\infty} 5\cdot t^{4}dt – \int_{-\infty}^{\frac{1}{2}} 5\cdot t^{4}dt =$$

$$=1 – \int_{-\infty}^{\frac{1}{2}} 5\cdot t^{4}dt= 1-\frac{5}{5} \cdot t^{5} \Big|_0^\frac{1}{2}=1-(\frac{1}{2})^5=0.96875$$

Second way, $$P(X > \frac{1}{2})=P(\frac{1}{2}<X < +\infty)$$ In this case $a=\frac{1}{2}$ and $b=+\infty$. Consequently, integral is equal to $$\int_{\frac{1}{2}}^{+\infty} 5\cdot t^{4}dt$$ Moreover, our function is zero outside of the interval $<0,1>$. Because of that we can rewrite our integral as $$\int_{\frac{1}{2}}^{1} 5\cdot t^{4}dt$$ Finally, we solve the given integral

$$\int_{\frac{1}{2}}^{1} 5\cdot t^{4}dt=\frac{5}{5} \cdot t^{5} \Big|_\frac{1}{2}^{1}=1-\Big(\frac{1}{2}\Big)^5=0.96875$$

**(c)** We’ve learned before that expected value is equal to $$E[X]:= \int_{-\infty}^{+\infty} tf(t)dt$$

We simply put our PDF and it’s edges into the given formula and solve the integral. It looks like this: $$E[X]:= \int_{0}^{1} t\cdot 5\cdot t^4 dt=5\cdot\int_{0}^{1}t^5 dt = \frac{5}{6}$$

Variance is calculated the same way as before, $Var(X)=E[X^{2}]-(E[X])^{2}$.

Since we already have $E[X]$, we need to find $E[X^{2}]$ and subtract them. $E[X^{2}]$ is calculated the same way as $E[X]$ but instead of putting $t\cdot f(t)$ in integral, we put $t^{2} f(t)$. Let’s do that. $$E[X^{2}]:= 5\int_{0}^{1} t^2 \cdot t^4 dt=5\cdot\int_{0}^{1}t^6 dt = \frac{5}{7}$$ Finally, $$Var(X)=E[X^{2}]-(E[X])^{2}=\frac{5}{7}-\Big(\frac{5}{6} \Big)^{2}=0.0198413$$

#### Uniform distribution

We’ve already mentioned probability distribution function, but lets extend our knowledge to continuous random variables.

**Proposition. Probability distribution function** also known as **cumulative distribution function (CDF)** of a random variable $X$ is $$F(x) =P(X \leq x), \quad \forall x \in \mathbb{R}$$

Specially, if $X$ is continuous with density $f$, its probability distribution function looks like this: $$F(x) =\int_{-\infty}^{x} f(t) dt, \quad \forall x \in \mathbb{R}$$

**Proposition.** Lets take a look at previous propositions and extend them in terms of probability distribution function.

**(a)** $$P(a < X \leq b) = P(X \leq b) – P(X \leq a)= F(b) – F(a)$$

**(b)** $$P(X>x) =1-P(X \leq x)=1-F(x)$$

**Proposition.** We say random variable $X$ has **uniform distribution** on interval $<a,b>$ if its distribution function looks like this $$f(t)=\begin{cases} \frac{1}{b-a}, & \text{if $a<t<b$}.\\ 0, & \text{otherwise}.\end{cases}$$ Additionally, we write it as $X \sim U(a,b)$

#### Normal distribution

Random variable $X$ has **normal** distribution with parameters $\mu \in \mathbb{R}$ and $\sigma^{2}>0$ if its distribution function looks like this $$f(t)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(t-\mu)^{2}}{2\sigma^{2}}},\quad \forall t \in \mathbb{R}$$ Additionally, we note it as $X \sim N(\mu, \sigma^{2})$.

**Proposition.**

**(a)** If $X \sim N(\mu, \sigma^{2})$, its expected value is $\mu$. Moreover, its variance is $\sigma^{2}$. Consequently, $\sigma=\sqrt{Var(X)}$is standard variance of $X$.

**(b)** For $X \sim N(\mu, \sigma^{2})$ we can define random variable $$Z:=\frac{X-\mu}{\sigma} \sim N(0,1)$$ Variable $Z\sim N(0,1)$ is called **standard normal** random variable and its distribution function is noted as $\Phi$. The graph of $\Phi$ looks like this:

Normal distribution is sometimes called **bell curve**.

The graph is symmetric around zero. Consequently, $\Phi(x)=1-\Phi(-x)$.

However, $\Phi$ can’t be explicitly calculated. We use the table below to find its value.

#### How to read the table?

Lets say we need to find $\Phi(1.14)$. Firstly, we separate the number $1.14$ into two parts:

- integer part and first decimal
- second decimal

**$1.1$$4$$=$$1.1$$+$$0.04$**

We look for the first summand in the table’s rows and for the second summand in the columns.

Consequently, $\Phi(1.14)=0.8729$.

#### Example 3

Let $X\sim N(1,4)$. Find $P(1 < X < \frac{3}{2})$.

**Solution**

From $X\sim N(1,4)$ we know that the expected value of $X$ is $\mu=1$ and its variance is $\sigma^{2}=4$. Using the previous proposition we can introduce standard normal random variable $Z$. $$Z\sim \frac{X-\mu}{\sigma}=\frac{X-1}{2}\sim N(0,1)$$ Be careful, we are given variance ($\sigma^{2}$), but for $Z$ we need standard deviation ($\sigma$). Now we want to extend the expression $P(1 < X < \frac{3}{2})$ to something like $P(a<Z<b)$. We do that by subtracting $1$ and dividing the whole expression inside of of brackets with $2$.$$P(1 < X < \frac{3}{2})=P(\frac{1-1}{2}<\frac{X-1}{2}<\frac{\frac{3}{2}-1}{2})=P(0<Z<\frac{1}{4})$$ In terms of $\Phi$ the above expression is equal to $$\Phi(\frac{1}{4})-\Phi(0)=\Phi(0.25)-\Phi(0.00)=0.5987-0.5=0.0987$$ The values of $\Phi(0.25)$ and $\Phi(0.00)$ were read from the table the same way we described.

#### Example 4

Lets say that the required time for a student to get to the college is distributed with expected value of $40$ minutes and standard deviation of $7$ minutes. His class starts at $12:15$.

**(a)** If he gets going at $11:40$, what is the probability he will make it on time.

**(b)** At what time should the student leave the house for the probability of making it in on time be at least $0.95$.

**Solution**

Let $X$ be random variable that represents the time of the travel. We were given the expected value and standard deviation:

$\bullet$ $\mu =40$ minutes

$\bullet$ $\sigma=7$ minutes

**(a)** For him to make it on time he needs to travel less than $35$ minutes. Consequently, we need $P(X<35)$. Once again, we extend it to normal standard random variable. $$P(\frac{X-40}{7}<\frac{35-40}{7})=P(\frac{X-40}{7}<-0.71429)$$ Now we can write this expression in terms of $\Phi$. $$P(\frac{X-40}{7}<-0.71429)=\Phi(-0.714)=1-\Phi(0.714)=1-0.7611=0.2389$$

The probability the student will make it to college on time is $23.89\%$.

**(b)** In this example we know the wanted probability, however we need to find the value of $X$ that gives us that probability. $$P(X\leq a)\geq 0.95$$ We extend the expressions similarly as before. $$P(\frac{X-40}{7}\leq \frac{a-40}{7})\geq 0.95$$ Further, $$\Phi(\frac{a-40}{7})\geq 0.95$$ In contrast to previous examples, we go the other way around. Firstly, we look for the first value of $Z$ greater than $0.95$ and then find the original $Z$ through right column and row. $$\Phi(1.65)\geq 0.95$$ Further, because $\Phi \geq 0.95$ for every number greater than $1.65$ (see the table), we have $$\frac{a-40}{7}\geq 1.65$$ When we solve this equation, we get $$a\geq 51.55$$ Finally, for the probability greater than $0.95$ students needs to start going to class at least $52$ min prior.