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[prob_dist] update according to editorial suggestions #408

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[prob_dist] update according to editorial suggestions
longye-tian Mar 21, 2024
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[prob_dist] Update on parenthesis
longye-tian Mar 21, 2024
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Update prob_dist.md
longye-tian Mar 21, 2024
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38 changes: 19 additions & 19 deletions lectures/prob_dist.md
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Expand Up @@ -86,7 +86,7 @@ Hence the second term takes all $x_i \leq x$ and sums their probabilities.

#### Uniform distribution

One simple example is the **uniform distribution**, where $p(x_i) = 1/n$ for all $n$.
One simple example is the **uniform distribution**, where $p(x_i) = 1/n$ for all $i$.

We can import the uniform distribution on $S = \{1, \ldots, n\}$ from SciPy like so:

Expand All @@ -96,7 +96,7 @@ u = scipy.stats.randint(1, n+1)
```


Here's the mean and variance
Here's the mean and variance:

```{code-cell} ipython3
u.mean(), u.var()
Expand All @@ -105,7 +105,7 @@ u.mean(), u.var()
The formula for the mean is $(n+1)/2$, and the formula for the variance is $(n^2 - 1)/12$.


Now let's evaluate the PMF
Now let's evaluate the PMF:

```{code-cell} ipython3
u.pmf(1)
Expand Down Expand Up @@ -140,7 +140,7 @@ plt.show()
```


The CDF jumps up by $p(x_i)$ and $x_i$.
The CDF jumps up by $p(x_i)$ at $x_i$.


```{exercise}
Expand Down Expand Up @@ -194,19 +194,19 @@ u.pmf(1)

#### Binomial distribution

Another useful (and more interesting) distribution is the **binomial distribution** on $S=\{0, \ldots, n\}$, which has PMF
Another useful (and more interesting) distribution is the **binomial distribution** on $S=\{0, \ldots, n\}$, which has PMF:

$$
p(i) = \binom{n}{i} \theta^i (1-\theta)^{n-i}
$$

Here $\theta \in [0,1]$ is a parameter.

The interpretation of $p(i)$ is: the number of successes in $n$ independent trials with success probability $\theta$.
The interpretation of $p(i)$ is: the probability of $i$ successes in $n$ independent trials with success probability $\theta$.

(If $\theta=0.5$, p(i) can be "how many heads in $n$ flips of a fair coin")
(If $\theta=0.5$, p(i) can be the probability of $i$ heads in $n$ flips of a fair coin.)
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Let's drop the outside parentheses and change this to "For example, if $\theta = 0.5$, then $p(i)$ is the probability of $i$ heads in $n$ flips of a fair coin."

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Dear John,

Got it. I have dropped the outside parentheses and changed the sentences.

Thank you for the suggestion!

Best,
Longye


The mean and variance are
The mean and variance are:

```{code-cell} ipython3
n = 10
Expand All @@ -220,7 +220,7 @@ u.mean(), u.var()

The formula for the mean is $n \theta$ and the formula for the variance is $n \theta (1-\theta)$.

Here's the PDF
Here's the PMF:

```{code-cell} ipython3
u.pmf(1)
Expand All @@ -236,7 +236,7 @@ plt.show()
```


Here's the CDF
Here's the CDF:

```{code-cell} ipython3
fig, ax = plt.subplots()
Expand All @@ -258,7 +258,7 @@ Using `u.pmf`, check that our definition of the CDF given above calculates the s
:class: dropdown
```

Here is one solution
Here is one solution:

```{code-cell} ipython3
fig, ax = plt.subplots()
Expand All @@ -283,9 +283,9 @@ $$
p(i) = \frac{\lambda^i}{i!} e^{-\lambda}
$$

The interpretation of $p(i)$ is: the number of events in a fixed time interval, where the events occur at a constant rate $\lambda$ and independently of each other.
The interpretation of $p(i)$ is: the probability of $i$ events in a fixed time interval, where the events occur at a constant rate $\lambda$ and independently of each other.

The mean and variance are
The mean and variance are:
```{code-cell} ipython3
λ = 2
u = scipy.stats.poisson(λ)
Expand All @@ -297,7 +297,7 @@ u.mean(), u.var()

The the expectation of Poisson distribution is $\lambda$ and the variance is also $\lambda$.

Here's the PMF
Here's the PMF:

```{code-cell} ipython3
λ = 2
Expand All @@ -321,7 +321,7 @@ plt.show()
### Continuous distributions


Continuous distributions are represented by a **density function**, which is a function $p$ over $\mathbb R$ (the set of all numbers) such that $p(x) \geq 0$ for all $x$ and
Continuous distributions are represented by a **probability density function**, which is a function $p$ over $\mathbb R$ (the set of all real numbers) such that $p(x) \geq 0$ for all $x$ and

$$ \int_{-\infty}^\infty p(x) dx = 1 $$

Expand Down Expand Up @@ -424,7 +424,7 @@ It has a nice interpretation: if $X$ is lognormally distributed, then $\log X$ i

It is often used to model variables that are "multiplicative" in nature, such as income or asset prices.

We can obtain the moments, PDF, and CDF of the normal density as follows:
We can obtain the moments, PDF, and CDF of the lognormal density as follows:

```{code-cell} ipython3
μ, σ = 0.0, 1.0
Expand Down Expand Up @@ -479,7 +479,7 @@ It is related to the Poisson distribution as it describes the distribution of th

It can be shown that, for this distribution, the mean is $1/\lambda$ and the variance is $1/\lambda^2$.

We can obtain the moments, PDF, and CDF of the normal density as follows:
We can obtain the moments, PDF, and CDF of the exponential density as follows:

```{code-cell} ipython3
λ = 1.0
Expand Down Expand Up @@ -535,7 +535,7 @@ This distribution has two parameters, $\alpha > 0$ and $\beta > 0$.
It can be shown that, for this distribution, the mean is $\alpha / (\alpha + \beta)$ and
the variance is $\alpha \beta / (\alpha + \beta)^2 (\alpha + \beta + 1)$.

We can obtain the moments, PDF, and CDF of the normal density as follows:
We can obtain the moments, PDF, and CDF of the Beta density as follows:

```{code-cell} ipython3
α, β = 3.0, 1.0
Expand Down Expand Up @@ -592,7 +592,7 @@ One interpretation is that if $X$ is gamma distributed and $\alpha$ is an
integer, then $X$ is the sum of $\alpha$ independent exponentially distributed
random variables with mean $1/\beta$.

We can obtain the moments, PDF, and CDF of the normal density as follows:
We can obtain the moments, PDF, and CDF of the Gamma density as follows:

```{code-cell} ipython3
α, β = 3.0, 2.0
Expand Down