Continuous uniform and normal distributions



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  • The following topics will be covered in this lecture:

    • Review of continuous distributions and their parameters
    • Continuous uniform distribution
    • Normal distribution
    • Standard normal distribution
    • Standardizing normal variables to compute z-scores

Review of main concepts

  • For a continuous random variable, the concepts from discrete random variables have direct analogs.

  • We have the following correspondences

Discrete Continuous
Probability mass function \( f(x) \) Probability density function \( f(x) \)
\( P(X=x_\alpha) = f(x_\alpha) \) \( P(a \leq X \leq b) = \int_{a}^b f(x)\mathrm{d}x \)
Cumulative distribution function \( F(x)=P(X\leq x) \) Cumulative distribution function \( F(X)=P(X\leq x) \)
\( F(x) = \sum_{x_\alpha \in \mathbf{R}} f(x_\alpha) \) \( F(x) = \int_{x\in \mathbb{R}}f(x)\mathrm{d}x \)
\( \mu = \sum_{x_\alpha \in \mathbf{R}} xf(x_\alpha) \) \( \mu = \int_{x\in \mathbb{R}} x f(x)\mathrm{d}x \)
\( \sigma^2 = \sum_{x_\alpha \in \mathbf{R}}(x - \mu)^2 f(x_\alpha) \) \( \sigma^2 = \int_{x\in \mathbb{R}}( x -\mu)^2 f(x)\mathrm{d}x \)
  • Due to the difference between discrete measurements and continuous measurements (where we can arbitrarily sub-divide units) the probability of measuring a single value of a continuous random variable always has probability zero.

  • Particularly, with continuous random variables, we always define probabilities over ranges of values, assuming some kind of truncation approximation.

  • Otherwise the ideas are extremely similar by replacing sums with integrals (or Riemann sums).

  • We will now consider two very common continuous probability distributions.

Continuous uniform disribution

  • Let's recall the copper current example from the last lecture.

  • The random variable \( X \) has a probability density defined as

    \[ \begin{align} f(x) = \begin{cases} 5.0 & x\in[4.9, 5.1]\\ 0.0 & \text{else} \end{cases} \end{align} \]

  • Similarly we found that the cumulative distribution function is given by

    \[ \begin{align} F(x)&= \begin{cases} 0.0 & x \in(-\infty, 4.9)\\ 5.0\left(x - 4.9\right) & x \in[4.9, 5.1]\\ 1.0 & x \in (5.1,\infty) \end{cases} \end{align} \]

  • The above is actually a specific example of the general continuous uniform probability distribution.

  • The continuous uniform is the probability model for a continuous random variable in which we assign equal probability to any sub-interval of equal length that lies in the range.

  • For example, in the above, we can see that

    \[ \begin{align} P(X\in [4.9,5.0]) = 0.5; & & P(X\in [5.0,5.1]) = 0.5. \end{align} \]

  • More generally, a simple \( \text{height}\times \text{width} \) area argument shows that for any \( a,b\in[4.9,5.1] \),

    \[ \begin{align} b-a=0.1 & & \Rightarrow & & P(X\in[a,b]) = 0.5. \end{align} \]

Continuous uniform disribution continued

  • More generally we have the general form of the continuous uniform distribution given as follows.
  • Let \( b>a \) be real numbers. A continuous random variable \( X \) with the probability density function \[ \begin{align} f(x) = \begin{cases} \frac{1}{b-a} & x \in[a,b]\\ 0 & \text{else} \end{cases}\end{align} \] is a continuous uniform random variable.
  • Notice that the above construction guarantees that the area under the density curve is exactly equal to one – the graph is pictured right.
Uniform distribution

Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

  • I.e., the area is compatible with our understanding of the total probability being equal to one: \[ \begin{align} \text{height}\times\text{width} = \frac{1}{b-a}\times \left(b-a\right) = 1 \end{align} \]
  • We often denote the range \( [a,b] \) as the support of the distribution, i.e., the range in which the density is non-zero.
  • Outside of the range \( [a,b] \), we say that there is no support for the probability as the probability of \( X \) lying outside of the range \( [a,b] \) is always zero.

Continuous uniform disribution continued

  • The simple form of the continuous uniform distribution makes it easy to compute some key parameters.

  • For example,

    \[ \begin{align} \mu &= \int_{-\infty}^\infty xf(x) \mathrm{d}x \\ &= \int_{a}^b \frac{x}{b-a} \mathrm{d}x \\ &= \frac{x^2}{2(b-a)}\big\vert_{a}^b\\ &= \frac{b^2 - a^2}{2(b-a)} \end{align} \]

  • But recall that we can factor

    \[ b^2 - a^2 = (b+a)(b-a) \] such that

    \[ \mu = \frac{b+a}{2}. \]

  • This corresponds once again to that the mean or expected value is the center of mass / average of the endpoints.

Continuous uniform disribution continued

  • Noting that \( \mu=\frac{b+a}{2} \), we can similarly obtain the variance as

    \[ \begin{align} \sigma^2 &= \int_{-\infty}^\infty (x - \mu)^2 f(x)\mathrm{d}x \\ &= \int_{a}^b \frac{\left(x - \frac{b+a}{2}\right)^2}{b-a}\mathrm{d}x \\ \end{align} \]

  • Recall, we can make a substitution as \( u = x - \frac{b+a}{2} \) such that \( du=dx \) and the range of integration is adjusted as

    \[ \begin{align} u_\text{lower} = a - \frac{b+a}{2} = \frac{a - b}{2}& & u_\text{upper} = b - \frac{b+a}{2} = \frac{b-a}{2} \end{align} \]

  • This results in,

    \[ \begin{align} \sigma^2 = \int_{\frac{a-b}{2}}^{\frac{b-a}{2}} \frac{u^2}{b-a}\mathrm{d}u = \frac{u^3}{3(b-a)}\big\vert_\frac{a-b}{2}^\frac{b-a}{2} \end{align} \]

  • Notice that the cubic term has the property that \( -(a-b)^3 = (b-a)^3 \), such that

    \[ \begin{align} \sigma^2 &= \frac{(b-a)^3}{8\times3(b-a)} - \frac{(a-b)^3}{8\times3(b-a)} = \frac{(b-a)^2}{12} \end{align} \]

Continuous uniform disribution continued

  • Finally, we can generally compute the cumulative distribution function of the continuous uniform as

    \[ \begin{align} F(x) &= \int_{a}^x \frac{1}{b-a} \mathrm{d}u \\ &= \frac{u}{b-a} \big\vert_{a}^x \\ &= \frac{x - a}{b-a} \end{align} \] where \( x \in[a,b] \).

  • Therefore, the general form is

    \[ \begin{align} F(x) = \begin{cases} 0 & x \in(-\infty, a) \\ \frac{x-a}{b-a} & x \in [a,b]\\ 1 & x \in [b,\infty) \end{cases} \end{align} \]

The normal distribution

Diagram of the percent of outcomes contained within each standard deviation of the mean
for a normal distribution.

Courtesy of Melikamp CC via Wikimedia Commons

  • The most widely used model for a continuous measurement is a normal random variable.
  • Whenever a random experiment is replicated, the average result over the replicates tends to have a normal distribution as the number of replicates becomes large.
  • De Moivre presented this fundamental result, known as the central limit theorem, in 1733.
  • Unfortunately this was lost for some time, and Gauss independently developed a normal distribution 100 years later.
  • Although De Moivre was later credited with the derivation, a normal distribution is also referred to as a Gaussian distribution.
  • The central limit theorem is the basis for classical statistical inference.
  • Using this, we can use this model to infer general properties of a population from representative samples.
  • For example, an automotive engineer may plan a study to average pull-off force measurements from several connectors.
  • If we assume that each measurement results from a replicate of a random experiment,
  • then, the normal distribution can be used to make approximate conclusions about the average pulloff force for all connectors of this design.

The normal distribution continued

  • Furthermore, sometimes the central limit theorem is less obvious.

  • For example, assume that the deviation (or error) in the length of a machined part is the sum of a large number of infinitesimal effects, e.g.,

    • such as temperature and humidity drifts,
    • vibrations,
    • cutting angle variations,
    • cutting tool wear,
    • bearing wear,
    • rotational speed variations,
    • mounting and fixture variations,
    • variations in numerous raw material characteristics, and
    • variations in levels of contamination.
  • If the component errors are independent and equally likely to be positive or negative, the total error can be shown to have an approximate normal distribution.

  • Furthermore, the normal distribution arises in the study of numerous basic physical phenomena.

  • For example, physicist James Maxwell developed a normal distribution from simple assumptions regarding the velocities of molecules.

The normal distribution continued

  • Unlike how we defined the density and used this to compute \( \mu \) and \( \sigma \) for the uniform, we will reverse this for the normal.
  • That is, we will use \( \mu \) and \( \sigma \) to define the density of the normal and parametrize the distribution.
  • Let us use the following notation for compactness where \[ \exp(x) = e^{x}. \]
  • Let the normal random variable \( X \) have mean \( \mu \) and standard deviation \( \sigma \). The probability density function is given as \[ \begin{align} f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{\left(x - \mu\right)^2}{2\sigma^2}\right) \end{align} \]
  • Recall how we considered \( \mu \) to be a measure of center and \( \sigma \) a measure of spread.
  • If we vary these two values, we can change the center of mass and the spread of the normal distribution:
Shapes of the normal density.

Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

  • A key point to note, however, is regardless of the shape, \( f(x)>0 \) for all \( x\in \mathbb{R} \).

The normal distribution continued

  • At the same time, while every interval has non-zero probability with the normal distribution;
Empirical rule for the normal density.

Courtesy of Montgomery & Runger, Applied Statistics and Probability for Engineers, 7th edition

  • the empirical rule tells us that the probability becomes extremely small beyond three standard deviations from the mean.
  • In particular:
    • \( \approx 68\% \) of the population lies within one standard deviation of the mean, \( [\mu - \sigma, \mu+\sigma] \);
    • \( \approx 95\% \) of the population lies within two standard deviations of the mean, \( [\mu - 2 \sigma , \mu + 2 \sigma] \)
    • \( \approx 99.7% \) of the population lies within three standard deviations of the mean, \( [\mu - 3 \sigma, \mu + 3 \sigma] \).
  • We can understand this by the form of the Gaussian density equation, where the term \[ \begin{align} \exp\left(-\frac{\left(x - \mu\right)^2}{2\sigma^2}\right) \end{align} \] shrinks at a super-exponential rate as \( x \) deviates from \( \mu \), relative to the size of \( \sigma \).
  • This has a limiting property where \[ \lim_{x \rightarrow \pm \infty} \exp\left(-\frac{\left(x - \mu\right)^2}{2\sigma^2}\right) = 0 \] but this is positive for any finite value \( x \).

The standard normal distribution

  • A special normal distribution that is commonly used for calculations is the standard normal.

  • The standard normal random variable \( Z \) has mean \( \mu=0 \) and standard deviation \( \sigma=1 \). The probability density function is therefore given as \[ \begin{align} f(z) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right). \end{align} \] We denote the cumulative distribution function of the standard normal \( \Phi(z)=P(Z\leq z) \).
  • The primary reason that the standard normal is often used in practice is because of the following result:

  • Let the normal random variable \( X \) have mean \( \mu \) and standard deviation \( \sigma \). The random variable defined \[ Z = \frac{X - \mu}{\sigma} \] follows the standard normal distribution above.
  • This says that for any normal random variable \( X \), if we shift the center to zero \( X - \mu \);

  • and re-scale the spread to one

    \[ Z=\frac{X - \mu}{\sigma}; \]

  • we can model the population with the easier-to-compute standard normal.

The standard normal distribution continued

  • The previous technique is known as standardizing a normal variable.

  • Because it is simple to standardize a variable, this technique was widely used before computers to calculate normal probabilities.

  • Specifically, if one has a normal model for the population \( X \) with mean \( \mu \) and \( \sigma \), this says,

    \[ \begin{align} X \leq x &&\Leftrightarrow && X - \mu \leq x - \mu && \Leftrightarrow && \frac{X - \mu}{\sigma} \leq \frac{x -\mu}{\sigma} & & Z \leq z \end{align} \]

  • Therefore,

    \[ \begin{align} P(X\leq x) = P(Z\leq z) \end{align} \] and this holds for all other ranges.

  • Note, this argument doesn't work for every distribution.

  • It is a special property of the normal that we can shift the center and scale the spread and the distribution remains normal.

  • This is one of several very special properties of this model.

The standard normal distribution continued

  • Because we standardize normal variables, the standard normal table for “z-scores” is a long used tool in statistics.

  • Rather than compute the probability for every normal variable individually, traditional statistics used a table for the standard normal to compute probabilities by standardizing the general normal variable.

  • Let \( x \) be a measurement of a normal random variable \( X \), which has mean \( \mu \) and standard deviation \( \sigma \). The z-score of the measurement is \[ \begin{align} z=\frac{x-\mu}{\sigma} \end{align} \] which measures how many standard deviations \( \sigma \) the measurement lies from the mean \( \mu \) in the positive or negative direction.
    • For our homework and the second midterm, we will practice the traditional approach using z-scores.

    • However, after the second midterm, we will focus on the modern approach using computer languages to make all such calculations.

    Computing z-scores from tables

    • To the right is a traditional z-score table.
    • In the rows, we can find a \( z \) value to the tenth decimal place.
    • In the columns, we can add values to the hundredth decimal place.
    • For example, if we want to compute the value \[ P(Z\leq 0.12) \] this value can be found precisely in first row and third column.
    • For an arbitrary value \( z \), we can thus compute the probability \[ P(Z\leq z) \] by finding a close approximation \( \tilde{z}\approx z \) up to rounding error and take the value in the table.
    • For an arbitrary range of values \( z_1 \leq Z \leq z_2 \), we can find \( \tilde{z}_1 \approx z_1 \) and \( \tilde{z}_2 \approx z_2 \) as above and find the probability as \[ \begin{align} P(z_1 \leq Z\leq z_2) &= P(Z\leq z_2) - P(Z\leq z_1) \\ &\approx P(Z\leq \tilde{z}_2) - P(Z\leq \tilde{z}_1) \end{align} \]
    • The calculation then just requires finding the values \( P(Z\leq \tilde{z}_2) \) and \( P(Z\leq \tilde{z}_1) \) in the appropriate row and column of the table.