Random Variables and Distribution Functions

 

📘 Random Variables – An Overview

🔹 1. Intuition and Example

A random variable (r.v.) is a real-valued function that assigns a number to each outcome of a random experiment.

Example:
Let the experiment be two coin tosses.
Sample Space S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}
Define a random variable XX = number of heads.

Outcome (ω)HH    HT    TH    TT
X(ω)2    1    1    0

🔹 2. Mathematical Definition

Let (S,B,P)(S, B, P) be a probability space:

  • SS: Sample space

  • BB: σ-field (collection of measurable subsets of SS)

  • PP: Probability function defined on BB

Then:

A random variable X(ω) is a real-valued function defined on SS, such that for every real number aa, the set

{ωS:X(ω)a}B\{\omega \in S : X(\omega) \leq a\} \in B

This ensures that probabilities like P(Xa)P(X \leq a) are well-defined.


🔹 3. Types of Random Variables

  • One-dimensional: Real-valued X:SRX: S \to \mathbb{R}

  • Two-dimensional: Vector-valued X:SR2X: S \to \mathbb{R}^2

  • n-dimensional: X:SRnX: S \to \mathbb{R}^n


🔹 4. Measurability

A random variable must be measurable, meaning it maps events in the sample space to real numbers in such a way that all standard probability statements (like P(Xa)P(X \leq a)) can be interpreted as events in the σ-field BB.


🔹 5. Notation and Probability Statements

  • XX: Random variable (uppercase letter)

  • xx: Value taken by XX (lowercase)

  • ω\omega: Sample point in SS

Some useful probability notations:

StatementMeaning
P(X=x)Probability that random variable XX equals xx
P(Xa)P(X \leq a)
Probability that XaX \leq a
P(a<Xb)P(a < X \leq b)
Probability that XX lies in the interval (a,b](a, b]
P(X=a or X=b)P(X = a \text{ or } X = b)
Union of events: P(X=a)+P(X=b)P(X = a) + P(X = b) if disjoint
P(X=a and X=b)P(X = a \text{ and } X = b)
Intersection of events (0 if aba \neq b)

🔹 6. Example Interpretation

If in an experiment, XX is the number of heads in two tosses:

P(X1)=P({HH,HT,TH})=34P(X \leq 1) = P(\{HH, HT, TH\}) = \frac{3}{4}

This is just the probability of all outcomes ω\omega such that X(ω)1X(\omega) \leq 1


🎲 Illustration 1: Tossing a Coin

✳️ Experiment: Toss a coin

  • Sample Space:

    S={ω1,ω2}={H,T}S = \{\omega_1, \omega_2\} = \{\text{H}, \text{T}\}

✳️ Define a Random Variable X(ω)X(\omega):

X(ω)={1,if ω=H (Head)0,if ω=T (Tail)X(\omega) = \begin{cases} 1, & \text{if } \omega = \text{H (Head)} \\ 0, & \text{if } \omega = \text{T (Tail)} \end{cases}

  • This is a Bernoulli random variable, since it only takes two values: 0 and 1.

  • It is an example of a discrete and finite-valued random variable.


🎲 Illustration 2: Rolling a Die

✳️ Experiment: Roll a fair six-sided die

  • Sample Space:

    S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}

✳️ Define Random Variable X(ω)X(\omega):

Let X(ω)=ωX(\omega) = \omega, i.e., the number appearing on the die.

So,

  • X(1)=1X(1) = 1

  • X(2)=2X(2) = 2

  • ...

  • X(6)=6X(6) = 6

✳️ Alternative Random Variable Y(ω):

Suppose we are only interested in whether the die roll is even or odd.

Then define:

Y(ω)={0,if ω is even (2,4,6)1,if ω is odd (1,3,5)Y(\omega) = \begin{cases} 0, & \text{if } \omega \text{ is even } (2, 4, 6) \\ 1, & \text{if } \omega \text{ is odd } (1, 3, 5) \end{cases}

So,

  • Y(2)=Y(4)=Y(6)=0Y(2) = Y(4) = Y(6) = 0

  • Y(1)=Y(3)=Y(5)=1Y(1) = Y(3) = Y(5) = 1

This again is a Bernoulli-type random variable since it takes only two values (0 and 1), representing a classification (even vs. odd).

📚 Theorems on Random Variables (without Proof)


🔸 Theorem 5.1 – Measurability

A function X(ω) from a sample space S to the real line R=(−∞,∞) is a random variable if and only if the set {ω∈S:X(ω)<a}

belongs to the σ-algebra B\mathcal{B} (i.e., is a measurable event) for all real numbers a.

🔸 Theorem 5.2 – Algebra of Random Variables

If X1 and X2​ are random variables and C is a constant, then:

CX1( scalar multiple)
CX1+CX2( sum)
X1X2(product)

are also random variables.

Remark:
It follows that any linear combination like C1X1+C2X2C_1X_1 + C_2X_2 is also a random variable.
In particular, X1X2X_1 - X_2 is a random variable.


🔸 Theorem 5.3 – Supremum, Infimum, Limits

If {Xn(ω),n≥1} is a sequence of random variables, then:

sup⁡nXn(ω),
inf⁡nXn(ω),
lim sup⁡n→∞Xn(ω),
lim inf⁡n→∞Xn(ω)

are all random variables, provided they are finite for all ω\omega.

🔸 Theorem 5.4 – Positive and Negative Parts, Absolute Value

If XX is a random variable, then the following are also random variables:

    1/X(ω), if X(ω)0X(\omega) \neq 0
    X+(ω)=max⁡(0,X(ω)) – positive part 
    X−(ω)=−min⁡(0,X(ω)) – negative part
    ∣X(ω)∣

🔸 Theorem 5.5 – Maximum and Minimum

If X1​ and X2​ are random variables, then:

max⁡(X1,X2)
min⁡(X1,X2)

are also random variables.

🔸 Theorem 5.6 – Function of a Random Variable


If X is a random variable and f(⋅) is a continuous function, then: f(X)is a random variable.

🔸 Theorem 5.7 – Monotonic Function

If X is a random variable and f(⋅) is an increasing function, then: f(X) is a random variable.

Corollary:
If ff is a function of bounded variation on every finite interval [a,b][a, b], and XX is a random variable, then f(X)f(X) is also a random variable.

📈 Distribution Function of a Random Variable

Let XX be a random variable defined on a probability space (S,B,P)(S, \mathcal{B}, P). The distribution function (also called the cumulative distribution function or cdf) of XX is denoted by:

FX(x)=P(Xx)=P(ωS:X(ω)x),<x<F_X(x) = P(X \leq x) = P\left( \omega \in S : X(\omega) \leq x \right), \quad -\infty < x < \infty

✅ Interpretation:

  • FX(x)F_X(x) gives the probability that the random variable XX takes a value less than or equal to xx.

  • It accumulates probabilities up to the value xx.


📌 Properties of a Distribution Function:

  1. Non-decreasing:
    If a<b, then FX(a)FX(b)F_X(a) \leq F_X(b)

  2. Right-continuous:

    limh0+FX(x+h)=FX(x)\lim_{h \to 0^+} F_X(x + h) = F_X(x)
  3. Limits:

    limxFX(x)=0,limxFX(x)=1\lim_{x \to -\infty} F_X(x) = 0,\quad \lim_{x \to \infty} F_X(x) = 1
  4. Bounded between 0 and 1:

    0FX(x)1
      5.Probability Between Two Points
  1. If a<ba < b, then:

    P(a<Xb)=F(b)F(a)P(a < X \leq b) = F(b) - F(a)

    📌 Explanation:
    The cdf accumulates the probability up to a value. So, the probability that XX lies between aa and bb (strictly greater than aa, less than or equal to bb) is just the difference in the cumulative probabilities at those two points.

📘 Discrete Random Variable

A discrete random variable (r.v.) is a random variable that takes at most a countable number of distinct values.

👉 In simpler terms:

  • A function defined on a discrete sample space (like outcomes of tossing a coin or rolling a die).

  • Values can be finite (e.g., 0, 1, 2) or countably infinite (like 1, 2, 3, ...).


🎯 Probability Mass Function (PMF)

If XX is a discrete random variable that takes values x1,x2,x3,x_1, x_2, x_3, \dots then the probability mass function (p.m.f.) is:

p(xi)=P(X=xi)p(x_i) = P(X = x_i)

✅ Conditions for a PMF:

  1. Non-negativity:

    p(xi)0for all ip(x_i) \geq 0 \quad \text{for all } i
  2. Total probability is 1:

    i=1p(xi)=1\sum_{i=1}^{\infty} p(x_i) = 1

📌 The collection {(xi,p(xi))}\{ (x_i, p(x_i)) \} is called the probability distribution of XX.


🧠 Remark:

  • The set of values taken by XX is called the spectrum or support of the random variable.

  • For any event ERE \subset \mathbb{R}, the probability:

    P(XE)=xiEp(xi)P(X \in E) = \sum_{x_i \in E} p(x_i)

✅ Example: Tossing a Fair Coin

Sample space:

S={H,T}S = \{ H, T \}

Define random variable XX as:

  • X(H)=1X(H) = 1

  • X(T)=0X(T) = 0

This means:

Outcome    Random Variable XX
            Probability
H            1                                    P(X=1)=12P(X = 1) = \frac{1}{2}
T            0P(X=0)=12
🧾 This is a Bernoulli random variable.


Example Problem:

A random variable X has the following probability function:
Values of   X.x :-2     -1     0         1         2         3
                p(x)   :0.1     k     0.2     2k      0.3      k
 Find the value of k. and calculate mean and variance.

Given Values of XX:

x=2, 1, 0, 1, 2, 3x = -2,\ -1,\ 0,\ 1,\ 2,\ 3

Corresponding probabilities p(x)p(x):

p(x)=0.1, k, 0.2, 2k, 0.3, kp(x) = 0.1,\ k,\ 0.2,\ 2k,\ 0.3,\ k

🔹 (i) Find the value of kk

Since the total probability must equal 1, we write:

0.1+k+0.2+2k+0.3+k=10.1 + k + 0.2 + 2k + 0.3 + k = 1

Combine like terms:

(0.1+0.2+0.3)+(k+2k+k)=10.6+4k=14k=10.6=0.4k=0.44=0.1(0.1 + 0.2 + 0.3) + (k + 2k + k) = 1 \Rightarrow 0.6 + 4k = 1 \Rightarrow 4k = 1 - 0.6 = 0.4 \Rightarrow k = \frac{0.4}{4} = 0.1

✅ So, k=0.1k = 0.1


🔹 Now calculate Mean E(X):

E(X)=xp(x)E(X) = \sum x \cdot p(x)
x        p(x)    x·p(x)
-2        0.1    -0.2
-1        0.1    -0.1
0        0.2        0
1        0.2         0.2
2        0.3     0.6
3        0.1     0.3



E(X)=0.20.1+0+0.2+0.6+0.3=0.8
E(X) = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8

Mean E(X)=0.8E(X) = 0.8


🔹 Now calculate Variance Var(X)\text{Var}(X):

We need:

Var(X)=E(X2)[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2

First calculate E(X2)=x2p(x)E(X^2) = \sum x^2 \cdot p(x)

x        p(x)    x²·p(x)
-2    4    0.1    0.4
-1    1    0.1    0.1
0    0    0.2    0
1    1    0.2    0.2
2    4    0.3    1.2
3    9    0.1    0.9
E(X2)=0.4+0.1+0+0.2+1.2+0.9=2.8E(X^2) = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 = 2.8
Var(X)=E(X2)[E(X)]2=2.8(0.8)2=2.80.64=2.16\text{Var}(X) = E(X^2) - [E(X)]^2 = 2.8 - (0.8)^2 = 2.8 - 0.64 = 2.16

Variance Var(X)=2.16\text{Var}(X) = 2.16


✅ Final Answers:

  • Value of kk: 0.1

  • Mean E(X)E(X): 0.8

  • Variance Var(X)\text{Var}(X): 2.16

Example-2

Given the probability function
x        =     0         1         2         3
p(x)   =     0·1     0·3     0·5     0·1
Let Y = X^2 + 2X , then find 
(i) the probability function of Y
(ii) mean and variance of Y


Given:

Random variable XX with the following probability function:

XX
        0        1        2        3
P(X)P(X)
        0.1        0.3        0.5        0.1
You are asked to define a new random variable:
Y=X2+2XY = X^2 + 2X

(i) Find the probability function of YY:

We'll compute YY for each value of XX:

Y=X2+2XY = X^2 + 2X
XX
     P(X)P(X)
Y=X2+2XY = X^2 + 2X
0     0.1      02+2×0=00^2 + 2×0 = 0
1    0.3     12+2×1=31^2 + 2×1 = 3
2    0.5     4+4=84 + 4 = 8
3    0.1     9+6=15
Now we define the probability function of  Y:
YY
      P(Y)P(Y)
0     0.1
3     0.3
8     0.5
15     0.1

(ii) Mean and Variance of YY:

🔹 Mean: E(Y)=YP(Y)E(Y) = \sum Y \cdot P(Y)

E(Y)=(0)(0.1)+(3)(0.3)+(8)(0.5)+(15)(0.1)=0+0.9+4.0+1.5=6.4E(Y) = (0)(0.1) + (3)(0.3) + (8)(0.5) + (15)(0.1) = 0 + 0.9 + 4.0 + 1.5 = 6.4

🔹 Variance:

Var(Y)=E(Y2)[E(Y)]2\text{Var}(Y) = E(Y^2) - [E(Y)]^2

We compute Y2P(Y)Y^2 \cdot P(Y):

YY
   Y2Y^2
Y2P(Y)Y^2 \cdot P(Y)
0   0     0×0.1=0.00 × 0.1 = 0.0
3   9     9×0.3=2.79 × 0.3 = 2.7
8   64   64×0.5=32.064 × 0.5 = 32.0
15  225 225×0.1=22.5225 × 0.1 = 22.5
E(Y2)=0+2.7+32.0+22.5=57.2E(Y^2) = 0 + 2.7 + 32.0 + 22.5 = 57.2
Var(Y)=57.2(6.4)2=57.240.96=16.24\text{Var}(Y) = 57.2 - (6.4)^2 = 57.2 - 40.96 = 16.24

Final Answers:

  • (i) Probability function of YY:

YY
   P(Y)
0  0.1
3  0.3
8  0.5
15   0.1
  • (ii) Mean of YY:

    E(Y)=6.4E(Y) = 6.4
  • Variance of YY:

    Var(Y)=16.24\text{Var}(Y) = 16.24

Continuous Random Variable

A random variable XX is said to be continuous if it can take any possible value within a certain interval. In other words:

A continuous random variable is one whose possible values cannot be listed individually, because they form an uncountably infinite set — such as all real numbers between two limits.

This means its values cannot be put into one-to-one correspondence with positive integers (unlike discrete variables).


Key Characteristics:

  • Can assume infinitely many values in a given range.

  • Is measurable to any desired level of precision.

  • The probability at any exact point is zero, i.e.,

    P(X=a)=0P(X = a) = 0
  • Instead, probability is defined over an interval, using a Probability Density Function (PDF):

    P(aXb)=abf(x)dxP(a \leq X \leq b) = \int_{a}^{b} f(x)\, dx

    where f(x)0f(x) \geq 0 and

    f(x)dx=1\int_{-\infty}^{\infty} f(x) \, dx = 1

Examples of Continuous Random Variables:

VariableDescription
Age        20.51 years, 20.512 years, etc.
Height        172.3 cm, 172.345 cm, etc.
Weight        64.2 kg, 64.2356 kg, etc.
Temperature        36.5°C, 36.5021°C, etc.
Time        10.5 seconds, 10.51 seconds, etc.

Comparison with Discrete Random Variables

FeatureDiscrete VariableContinuous Variable
Value type    Countable (1, 2, 3…)    Uncountable (real values)
Probability distribution    Probability Mass Function (PMF)    Probability Density Function (PDF)
Example    Number of students, toss of a die    Height, weight, age
Probability at exact value    >0> 0
    =0= 0

Probability Density Function (PDF) – Concept and Definition

A Probability Density Function (PDF) is used to describe the probability distribution of a continuous random variable.

Let XX be a continuous random variable, and let f(x)f(x) be a continuous function defined over the domain of XX. Then:

P(xXx+dx)=f(x)dxP(x \leq X \leq x + dx) = f(x) \cdot dx

This means that the probability that the random variable XX takes a value within the small interval (x,x+dx)(x, x + dx) is approximately:

Probabilityf(x)dx\text{Probability} \approx f(x) \, dx

Here,

  • f(x)f(x) is the probability density function,

  • dxdx is a very small interval (infinitesimally small),

  • f(x)dxf(x) \cdot dx gives the probability that XX lies in the interval (x,x+dx)(x, x + dx)


Key Properties of PDF

  1. Non-Negativity:

    f(x)0for all real xf(x) \geq 0 \quad \text{for all real } x
  2. Total Area Under the Curve Is 1:

    f(x)dx=1\int_{-\infty}^{\infty} f(x)\, dx = 1

    This expresses that the total probability is 1.

  3. Probability over an interval:

    P(aXb)=abf(x)dxP(a \leq X \leq b) = \int_{a}^{b} f(x) \, dx
  4. Probability at an exact value is zero:

    P(X=x)=0P(X = x) = 0

    Because the width of the interval is zero, the probability is zero.


Example

If the PDF of a continuous random variable XX is:

f(x)={2xif 0x10otherwisef(x) = \begin{cases} 2x & \text{if } 0 \leq x \leq 1 \\ 0 & \text{otherwise} \end{cases}

Then,

  • The probability that XX lies between 0.2 and 0.5 is:

    P(0.2X0.5)=0.20.52xdx=[x2]0.20.5=0.250.04=0.21P(0.2 \leq X \leq 0.5) = \int_{0.2}^{0.5} 2x \, dx = [x^2]_{0.2}^{0.5} = 0.25 - 0.04 = 0.21

Here's the graph of the probability density function (PDF) f(x)=2xf(x) = 2x for 0x10 \leq x \leq 1

The blue curve represents the function f(x)=2x

The orange shaded area shows the probability that the random variable X lies in the interval [0.2,0.5], which we calculated earlier as 0.21.

📌 Important Remark: Probability at a Point for Continuous Random Variables

✅ For Discrete Random Variables:

  • A discrete random variable XX can take on specific values like x=1,2,3, etc.

  • The probability at a particular point is not zero:

    P(X=c)0for some cP(X = c) \neq 0 \quad \text{for some } c
  • For example, if P(X=2)=0.4, that means there's a 40% chance that X=2X = 2.


✅ For Continuous Random Variables:

  • A continuous random variable can take infinitely many values over an interval (like time, height, weight).

  • Probability at a single point is always zero:

    P(X=c)=0for all cP(X = c) = 0 \quad \text{for all } c

    This is because the total probability (area under the curve) is spread over an interval, and the area at a single point is zero.


🔁 This Leads to an Important Equality:

For continuous random variables, whether you include or exclude the endpoints of an interval does not matter:

P(aXb)=P(a<Xb)=P(aX<b)=P(a<X<b)P(a \leq X \leq b) = P(a < X \leq b) = P(a \leq X < b) = P(a < X < b)

This is valid because:

P(X=a)=0andP(X=b)=0P(X = a) = 0 \quad \text{and} \quad P(X = b) = 0

So, interval types (open, closed, half-open) are interchangeable for probability purposes in continuous distributions.

📘 Measures for Continuous Probability Distributions

Let f(x)f(x) be a probability density function (p.d.f.) defined on the interval [a,b][a, b].

1. Arithmetic Mean (Expected Value)

μ=abxf(x)dx\mu = \int_a^b x \cdot f(x)\, dx


2. Harmonic Mean (H)

1H=ab1xf(x)dx\frac{1}{H} = \int_a^b \frac{1}{x} f(x)\, dx


3. Geometric Mean (G)

logG=ablogxf(x)dxG=elogxf(x)dx\log G = \int_a^b \log x \cdot f(x)\, dx \quad \Rightarrow \quad G = e^{\int \log x \cdot f(x)\, dx}


4. Moments

(a) r-th moment about origin:

μr=abxrf(x)dx\mu_r' = \int_a^b x^r \cdot f(x)\, dx

(b) r-th moment about a point AA:

μr(A)=ab(xA)rf(x)dx\mu_r(A) = \int_a^b (x - A)^r \cdot f(x)\, dx

(c) Central moments (about mean):

μr=ab(xμ)rf(x)dx\mu_r = \int_a^b (x - \mu)^r \cdot f(x)\, dx

Specifically:

  • Variance:

σ2=μ2=ab(xμ)2f(x)dx=μ2μ2\sigma^2 = \mu_2 = \int_a^b (x - \mu)^2 \cdot f(x)\, dx = \mu_2' - \mu^2


5. Skewness and Kurtosis

  • Skewness γ1=μ3σ3\gamma_1 = \frac{\mu_3}{\sigma^3}

  • Kurtosis γ2=μ4σ4\gamma_2 = \frac{\mu_4}{\sigma^4}

You can compute μ3\mu_3, μ4\mu_4 using:

μ3=μ33μ1μ2+2μ13\mu_3 = \mu_3' - 3\mu_1'\mu_2' + 2\mu_1'^3 μ4=μ44μ3μ1+6μ2μ123μ14\mu_4 = \mu_4' - 4\mu_3'\mu_1' + 6\mu_2'\mu_1'^2 - 3\mu_1'^4


6. Median (M)

The value of MM satisfies:

aMf(x)dx=12\int_a^M f(x)\, dx = \frac{1}{2}


7. Mean Deviation (about mean μ\mu)

M.D.=abxμf(x)dx\text{M.D.} = \int_a^b |x - \mu| \cdot f(x)\, dx


8. Quartiles and Deciles

  • First Quartile Q1Q_1:

aQ1f(x)dx=14\int_a^{Q_1} f(x)\, dx = \frac{1}{4}

  • Third Quartile Q3Q_3:

aQ3f(x)dx=34\int_a^{Q_3} f(x)\, dx = \frac{3}{4}

  • i-th Decile DiD_i:

aDif(x)dx=i10\int_a^{D_i} f(x)\, dx = \frac{i}{10}


9. Mode

  • The value of xx where the density f(x)f(x) is maximum:

f(x)=0andf(x)<0f'(x) = 0 \quad \text{and} \quad f''(x) < 0

Example  Problem-1

  • A continuous random variable XX with p.d.f.

    f(x)=A+Bxfor 0x1f(x) = A + Bx \quad \text{for } 0 \leq x \leq 1
  • The mean of the distribution is 1.

find constants AA and BB.


✅ Step 1: Use the property that total probability = 1

01f(x)dx=101(A+Bx)dx=1\int_0^1 f(x)\, dx = 1 \Rightarrow \int_0^1 (A + Bx)\, dx = 1
=[Ax+B2x2]01=A+B2=1(Equation 1)= \left[Ax + \frac{B}{2}x^2\right]_0^1 = A + \frac{B}{2} = 1 \quad \text{(Equation 1)}


✅ Step 2: Use the mean formula

Mean=01xf(x)dx=101x(A+Bx)dx=1\text{Mean} = \int_0^1 x f(x)\, dx = 1 \Rightarrow \int_0^1 x(A + Bx)\, dx = 1
=01(Ax+Bx2)dx=[A2x2+B3x3]01=A2+B3=1(Equation 2)= \int_0^1 (Ax + Bx^2)\, dx = \left[\frac{A}{2}x^2 + \frac{B}{3}x^3\right]_0^1 = \frac{A}{2} + \frac{B}{3} = 1 \quad \text{(Equation 2)}


✅ Step 3: Solve the two equations

From Equation 1:

A+B2=1A=1B2(3)A + \frac{B}{2} = 1 \Rightarrow A = 1 - \frac{B}{2} \tag{3}

Substitute into Equation 2:

1B22+B3=112B4+B3=1\frac{1 - \frac{B}{2}}{2} + \frac{B}{3} = 1 \Rightarrow \frac{1}{2} - \frac{B}{4} + \frac{B}{3} = 1
B4+B3=12B(14+13)=12B(112)=12B=6\Rightarrow -\frac{B}{4} + \frac{B}{3} = \frac{1}{2} \Rightarrow B\left(-\frac{1}{4} + \frac{1}{3}\right) = \frac{1}{2} \Rightarrow B\left(\frac{1}{12}\right) = \frac{1}{2} \Rightarrow B = 6

Substitute back into Equation (3):

A=162=13=2A = 1 - \frac{6}{2} = 1 - 3 = -2


✅ Final Answer:

A=2,B=6\boxed{A = -2,\quad B = 6}

Example Problem-2:

We are given a probability density function (PDF):

f(x)=cx2(1x),for 0<x<1f(x) = c x^2 (1 - x), \quad \text{for } 0 < x < 1

We are to find c


(i) The constant cc

We know:

01f(x)dx=101cx2(1x)dx=1\int_0^1 f(x) \, dx = 1 \Rightarrow \int_0^1 c x^2 (1 - x) \, dx = 1

Factor out cc:

c01x2(1x)dx=1c \int_0^1 x^2(1 - x) \, dx = 1

First, expand the integrand:

x2(1x)=x2x3x^2(1 - x) = x^2 - x^3

Now integrate:

c01(x2x3)dx=c[x33x44]01=c(1314)=c(4312)=c112c \int_0^1 (x^2 - x^3)\, dx = c \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = c\left(\frac{1}{3} - \frac{1}{4}\right) = c\left(\frac{4 - 3}{12}\right) = c \cdot \frac{1}{12}

Set equal to 1:

c12=1c=12\frac{c}{12} = 1 \Rightarrow \boxed{c = 12}


(ii) Mean μ=E[X]\mu = E[X]

E[X]=01xf(x)dx=01x12x2(1x)dx=1201x3(1x)dxE[X] = \int_0^1 x f(x)\, dx = \int_0^1 x \cdot 12x^2(1 - x) \, dx = 12 \int_0^1 x^3(1 - x) \, dx

Expand the integrand:

x3(1x)=x3x4x^3(1 - x) = x^3 - x^4

Integrate:

12[x44x55]01=12(1415)=12120=3512 \left[\frac{x^4}{4} - \frac{x^5}{5}\right]_0^1 = 12\left(\frac{1}{4} - \frac{1}{5}\right) = 12 \cdot \frac{1}{20} = \boxed{\frac{3}{5}}


(iii) Variance σ2=E[X2](E[X])2\sigma^2 = E[X^2] - (E[X])^2

First compute E[X2]:

E[X2]=01x2f(x)dx=01x212x2(1x)dx=1201x4(1x)dxE[X^2] = \int_0^1 x^2 f(x)\, dx = \int_0^1 x^2 \cdot 12x^2(1 - x) \, dx = 12 \int_0^1 x^4(1 - x)\, dx
=1201(x4x5)dx=12[x55x66]01=12(1516)=12130=25= 12 \int_0^1 (x^4 - x^5)\, dx = 12 \left[\frac{x^5}{5} - \frac{x^6}{6}\right]_0^1 = 12 \left(\frac{1}{5} - \frac{1}{6}\right) = 12 \cdot \frac{1}{30} = \frac{2}{5}

Now, use the formula:

Var(X)=E[X2](E[X])2=25(35)2=25925=10925=125\text{Var}(X) = E[X^2] - (E[X])^2 = \frac{2}{5} - \left(\frac{3}{5}\right)^2 = \frac{2}{5} - \frac{9}{25} = \frac{10 - 9}{25} = \boxed{\frac{1}{25}}


✅ Final Answers:

  • (i) c=12​

  • (ii) Mean μ=35\mu = \boxed{\frac{3}{5}}

  • (iii) Variance σ2=125\sigma^2 = \boxed{\frac{1}{25}}

Comments

Popular posts from this blog

GNEST305 Introduction to Artificial Intelligence and Data Science KTU BTech S3 2024 Scheme - Dr Binu V P

Basics of Machine Learning

Types of Machine Learning Systems