三个指数分布观测值的中位数X的PDF求解及思路问询

Hey there! Your instinct to start with the CDF and then differentiate to get the PDF is exactly the right approach—let's walk through this step by step to clear up where you might have gotten stuck.

Background Setup #

First, let's formalize the problem: assume ( Y_1, Y_2, Y_3 ) are independent and identically distributed (i.i.d.) exponential random variables with rate parameter ( \lambda ). The CDF of a single ( Y_i ) is:

F_Y(y) = 
\begin{cases}
1 - e^{-\lambda y} & y \geq 0, \\
0 & y < 0.
\end{cases}

We need to find the PDF of ( X = \text{Median}(Y_1, Y_2, Y_3) ).

Step 1: Translate the Median Event to a Probability Statement #

The median of three variables is the second smallest value when sorted. So ( X \leq x ) if and only if at least two of the ( Y_i ) are less than or equal to ( x ). Why? Because if the middle value is ≤x, at least two values must be ≤x (the middle one and at least one more). Conversely, if at least two values are ≤x, the middle value will definitely be ≤x.

This gives us the CDF of ( X ):

F_X(x) = P(X \leq x) = P(\text{at least 2 of } Y_1,Y_2,Y_3 \leq x)

Step 2: Calculate the CDF Using Combinatorics #

We can break this probability into two mutually exclusive cases: exactly 2 variables ≤x, or all 3 variables ≤x. Using the binomial probability formula (since each ( Y_i ) is independent):

F_X(x) = \binom{3}{2} \left[ F_Y(x) \right]^2 \left[ 1 - F_Y(x) \right] + \binom{3}{3} \left[ F_Y(x) \right]^3

Where ( \binom{n}{k} ) is the binomial coefficient (number of ways to choose k items from n).

Step 3: Substitute the Exponential CDF and Simplify #

Let's plug in ( F_Y(x) = 1 - e^{-\lambda x} ):

1. First term (exactly 2 variables ≤x):

   \binom{3}{2} (1 - e^{-\lambda x})^2 e^{-\lambda x} = 3 \left(1 - 2e^{-\lambda x} + e^{-2\lambda x}\right) e^{-\lambda x} = 3e^{-\lambda x} - 6e^{-2\lambda x} + 3e^{-3\lambda x}
   

2. Second term (all 3 variables ≤x):

   \binom{3}{3} (1 - e^{-\lambda x})^3 = (1 - 3e^{-\lambda x} + 3e^{-2\lambda x} - e^{-3\lambda x})
   

Add these two terms together and simplify:

F_X(x) = \left(3e^{-\lambda x} - 6e^{-2\lambda x} + 3e^{-3\lambda x}\right) + \left(1 - 3e^{-\lambda x} + 3e^{-2\lambda x} - e^{-3\lambda x}\right)

After canceling like terms:

F_X(x) = 
\begin{cases}
1 - 3e^{-2\lambda x} + 2e^{-3\lambda x} & x \geq 0, \\
0 & x < 0.
\end{cases}

Step 4: Differentiate the CDF to Get the PDF #

To find the PDF ( f_X(x) ), take the derivative of ( F_X(x) ) with respect to ( x ):

f_X(x) = \frac{d}{dx} F_X(x) = \frac{d}{dx} \left(1 - 3e^{-2\lambda x} + 2e^{-3\lambda x}\right)

Calculating the derivative:

f_X(x) = 
\begin{cases}
6\lambda e^{-2\lambda x} - 6\lambda e^{-3\lambda x} & x \geq 0, \\
0 & x < 0.
\end{cases}

We can factor this to make it cleaner:

f_X(x) = 6\lambda \left(e^{-2\lambda x} - e^{-3\lambda x}\right) \quad \text{for } x \geq 0

Step 5: Verify the PDF is Valid #

A quick check to ensure this is a valid PDF: integrate it from 0 to ∞ and confirm the result is 1:

\int_0^\infty 6\lambda \left(e^{-2\lambda x} - e^{-3\lambda x}\right) dx = 6\lambda \left[ -\frac{1}{2\lambda}e^{-2\lambda x} + \frac{1}{3\lambda}e^{-3\lambda x} \right]_0^\infty = 6\lambda \left(0 - \left(-\frac{1}{2\lambda} + \frac{1}{3\lambda}\right)\right) = 6\lambda \cdot \frac{1}{6\lambda} = 1

Perfect, that checks out!

Key Takeaways #

• The core insight is translating the median event into a countable number of binomial cases (at least two variables ≤x). This is a common strategy for order statistics (which the median is, for n=3).
• Always verify your PDF by checking its integral over the entire support equals 1—it's a quick way to catch algebraic mistakes.

内容的提问来源于stack exchange，提问作者Ayman