Probability distributions in MATLAB - Time & Space Complexity

Time Complexity: Probability distributions

O(n)

Understanding Time Complexity

When working with probability distributions in MATLAB, it's important to understand how the time needed to compute values grows as the input size increases.

We want to know how the execution time changes when we calculate probabilities for many data points.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

% Calculate normal distribution PDF for each value in data array
data = linspace(-5, 5, n);
pdf_values = zeros(1, n);
for i = 1:n
    pdf_values(i) = (1 / sqrt(2 * pi)) * exp(-0.5 * data(i)^2);
end
    

This code computes the probability density function (PDF) of a normal distribution for n points stored in the array data.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that calculates the PDF value for each element in the data array.
• How many times: The loop runs exactly n times, once for each data point.

How Execution Grows With Input

As the number of points n increases, the total calculations increase in a straight line.

Input Size (n)	Approx. Operations
10	10 calculations
100	100 calculations
1000	1000 calculations

Pattern observation: Doubling the input size roughly doubles the work done.

Final Time Complexity

Time Complexity: O(n)

This means the time to compute grows directly in proportion to the number of data points.

Common Mistake

[X] Wrong: "Calculating the PDF for many points is constant time because each calculation is simple."

[OK] Correct: Even though one calculation is quick, doing it many times adds up, so total time grows with the number of points.

Interview Connect

Understanding how time grows with input size helps you write efficient code when working with data and probability, a common task in many projects.

Self-Check

"What if we used MATLAB's vectorized operations instead of a for-loop? How would the time complexity change?"

Practice

1. Which MATLAB function is used to create a probability distribution object?

easy

A. rand

B. makedist

C. histogram

D. plot

Solution

1. Step 1: Understand the purpose of each function

   makedist creates a probability distribution object, while rand generates random numbers, histogram plots data distribution, and plot creates graphs.

2. Step 2: Identify the correct function for creating distributions

   Only makedist is designed to create probability distribution objects in MATLAB.

3. Final Answer:

   makedist -> Option B

4. Quick Check:

   Function to create distribution = makedist [OK]

Hint: Use makedist to create distributions in MATLAB [OK]

Common Mistakes:

• Confusing makedist with rand
• Using histogram to create distributions
• Trying to use plot for distributions

2. Which of the following is the correct syntax to create a normal distribution with mean 0 and standard deviation 1 in MATLAB?

easy

A. pd = makedist('normal', 0, 1);

B. pd = makedist('Normal', 'mean', 0, 'std', 1);

C. pd = makedist('Normal', 0, 1);

D. pd = makedist('Normal', 'mu', 0, 'sigma', 1);

Solution

1. Step 1: Recall the correct parameter names for normal distribution

   In MATLAB, the normal distribution parameters are specified as 'mu' for mean and 'sigma' for standard deviation.

2. Step 2: Check the syntax for makedist

   The correct syntax uses name-value pairs: makedist('Normal', 'mu', 0, 'sigma', 1). Other options use incorrect parameter names or formats.

3. Final Answer:

   pd = makedist('Normal', 'mu', 0, 'sigma', 1); -> Option D

4. Quick Check:

   Normal dist params = 'mu' and 'sigma' [OK]

Hint: Use 'mu' and 'sigma' as parameter names for normal dist [OK]

Common Mistakes:

• Using lowercase 'normal' instead of 'Normal'
• Using 'mean' and 'std' instead of 'mu' and 'sigma'
• Passing parameters without name-value pairs

3. What is the output of the following MATLAB code?

pd = makedist('Exponential', 'mu', 2);
p = pdf(pd, 1);
disp(round(p, 3));

medium

A. 0.303

B. 0.368

C. 0.500

D. 0.607

Solution

1. Step 1: Understand the exponential distribution PDF formula

   The PDF for exponential with mean mu=2 is f(x) = (1/2)*exp(-x/2). For x=1, f(1) = 0.5 * exp(-0.5) ≈ 0.5 * 0.6065 = 0.3033.

2. Step 2: Compute and round the value

   p ≈ 0.3033, round(p, 3) = 0.303.

3. Final Answer:

   0.303 -> Option A

4. Quick Check:

   (1/mu)*exp(-x/mu) at x=1, mu=2 ≈ 0.303 [OK]

Hint: Recall pdf = (1/mu)*exp(-x/mu) for exponential [OK]

Common Mistakes:

• Ignoring the (1/mu) factor
• Confusing PDF with CDF
• Rounding errors causing mismatch

4. Identify the error in this MATLAB code snippet:

pd = makedist('Binomial', 'n', 10, 'p', 0.5);
prob = cdf(pd, 11);

medium

A. The probability p must be an integer

B. makedist does not support 'Binomial' distribution

C. The cdf input value 11 exceeds the number of trials n

D. The variable name 'prob' is reserved

Solution

1. Step 1: Understand binomial distribution parameters

   Binomial distribution with n=10 means possible outcomes are from 0 to 10.

2. Step 2: Check the cdf input value

   Calling cdf(pd, 11) asks for probability of outcomes ≤ 11, but 11 is outside the valid range (max 10).

3. Step 3: Identify the error

   Input to cdf must be within 0 to n. Using 11 is invalid and causes an error or unexpected result.

4. Final Answer:

   The cdf input value 11 exceeds the number of trials n -> Option C

5. Quick Check:

   cdf input must be ≤ n for binomial [OK]

Hint: cdf input must be within distribution support range [OK]

Common Mistakes:

• Using cdf input outside valid range
• Thinking p must be integer
• Assuming makedist lacks binomial support

5. You want to calculate the probability that a normally distributed variable with mean 5 and standard deviation 2 is less than 7 in MATLAB using a probability distribution object. Which code correctly computes this?

hard

A. pd = makedist('Normal', 'mu', 5, 'sigma', 2); p = cdf(pd, 7);

B. p = normcdf(7, 5, 2);

C. pd = makedist('Normal', 5, 2); p = pdf(pd, 7);

D. p = cdf('Normal', 7, 5, 2);

Solution

1. Step 1: Understand how to create a normal distribution object

   Use makedist('Normal', 'mu', 5, 'sigma', 2) to create the distribution with mean 5 and std 2.

2. Step 2: Use cdf to find probability less than 7

   Calling cdf(pd, 7) returns the cumulative probability up to 7.

3. Step 3: Check other options

   p = normcdf(7, 5, 2); uses normcdf which is valid but not using makedist. pd = makedist('Normal', 5, 2); p = pdf(pd, 7); incorrectly uses pdf instead of cdf. p = cdf('Normal', 7, 5, 2); has wrong syntax for cdf.

4. Final Answer:

   pd = makedist('Normal', 'mu', 5, 'sigma', 2); p = cdf(pd, 7); -> Option A

5. Quick Check:

   Use makedist + cdf for cumulative probability [OK]

Hint: Use cdf with makedist object for cumulative probability [OK]

Common Mistakes:

• Using pdf instead of cdf
• Wrong syntax for cdf
• Ignoring distribution parameters