Continuous Random Variable

Continuous random variables have an inifinite number of values for any given interval. While similar, the approach to analysis is very different from discrete variables

• Summation becomes integration
• Probability becomes area under a curve

Probability Distribution Function $f(x)$

The probability density function (pdf) maps a continuous variable to a probability density.

As the name "density" suggests, the area under the pdf curve between a range is the probability of the variable being in that range.

$P(c\leq x\leq d)=\int _{c}^{d}f(x)dx=F(d)-F(c)$

Total area under the curve must be $1$, as chances of events happening is 100% if the range includes all possible events.

$\int _{-\infty }^{\infty }f(x)dx=1$

There is no area under a single point

$P(X=a)=0$

Mean and Variance

The mean and variance calculations are pretty much the same as that of discrete random variables, except the summations are swapped out for integrals.

$E(X)=\mu _{X}=\int _{-\infty }^{\infty }xf(x)dx$

$Var(X)=\sigma _{X}^{2}=\int _{-\infty }^{\infty }(x-\mu _{X})^{2}f(x)dx=\int _{-\infty }^{\infty }x^{2}f(x)dx-\mu _{X}^{2}$

Median and Percentile

The a-th percentileis the point at which a percent the area under the curve is to one side. You want $P(X\leq x)$ to be a%, the calculation of which is in the page above.

By the same logic, the quartiles are at 25%, 50%, and 75% accordingly.

Uniform Distribution $X\sim Uniform(a,b)$

Uniform random variable is described by two parameters: $a$ is minimum, and $b$ is maximum. It has a rectangular distribution, where every point has the same probability density.

PDF

$f(x)={\begin{cases}{\frac {1}{b-a}}&a\leq x\leq b\\0&{\text{otherwise}}\end{cases}}$

CDF

$F(x)={\begin{cases}0&x<a\\{\frac {x-a}{b-a}}&a\leq x\leq b\\1&x>b\end{cases}}$

Mean

$\mu _{X}={\frac {a+b}{2}}$

Variance

$\sigma ^{2}={\frac {1}{12}}(b-a)^{2}$

Exponential Distribution $X\sim Exp(\lambda )$

The exponential distribution models events that occurs

• Continuously
• Independently
• At a constant average rate

It takes in one parameter: $\lambda$, the rate parameter. It is defined by the mean below.

Exponential distribution has the memoryless property: the probability to an event does not change no matter how much time has passed.

In probability terms, the probability that we must wait an additional $t$ units given that we have waited $s$ units

$P(T>t+s|T>s)=P(T>t)=e^{-\lambda t}$

Notably, it models time until some event has happened.

PDF

$f(x)={\begin{cases}\lambda e^{-\lambda x}&a\leq x\leq b\\0&{\text{otherwise}}\end{cases}}$

CDF

$F(x)=1-e^{-\lambda x}$

Mean

Integration by parts

$\mu _{X}={\frac {1}{\lambda }}$

Variance

Integration by parts

$\sigma ^{2}={\frac {1}{\lambda ^{2}}}$

Exponential and Poisson

Exponential distribution and poisson RVs are related:

• $X\sim Poisson(\lambda )$: the number of events in a unit time
• $X\sim Exp(\lambda )$: waiting time until an event

Normal Random Variable

Normal random variables are the most widly used continuous RV in statistics, characterizing many natural phenomenons. It is the famous bell curve.

They are characterized by two parameters: mean and variance.

$Y\sim N(\mu _{Y},\sigma _{Y}^{2})$

Normal random variables are perfectly symmetric at the mean.

Standardizing Normal Distribution

Standardization of a data means to make its mean 0 and its standard deviation 1. We do this by subtracting the mean and dividing by the standard deviation:

$Z={\frac {Y-\mu }{\sigma }}$

Intuitively, this moves the dataset and changes the scale. We do this to simplify probability calculations.

Z score

The z-score is the number of standard deviations above or below the mean. A positive z score is above, and a negative is below.

$z={\frac {y-\mu }{\sigma }}$

PDF

The pdf for normal random variable is the following.

$f(y)={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {1}{2}}{\frac {(y-\mu )^{2}}{\sigma ^{2}}}}$

After standardizing the normal RV, we can use the following instead.

$f(y)={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {1}{2}}z^{2}}$

where $z$ is the z-score covered in the last section.

$f(z)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}z^{2}}$

CDF

$F(y)=1-e^{-\lambda y}$