ENPR300 Probability & Random Process Assessment 2026 | University of Dubai

This instrument assesses the following Course Learning Outcomes (CLO):

(1) Let {Xᵢ}ᵢ₌₁^∞ be a sequence of i.i.d. Bernoulli random variables with

P(Xᵢ = 1) = 0.3, P(Xᵢ = 0) = 0.7.        Q1 [22 pts] (1) Bernoulli Variables

A. [3pts] For Bernoulli

Expected Value:

E[Xᵢ] = 1×P(Xᵢ=1) + 0×P(Xᵢ=0)
      = 1×0.3 + 0×0.7
      E[Xᵢ] = 0.3

Variance:

σ(i) = (1 − )

σ(i) = 0.3 × 0.7

σ(i) = 0.21

B. [3pts] By Strong Law of Large Numbers,

→ μ = E[X] = 0.3        as n → ∞

→ 0.3 as n → ∞

C. [4pts] No contradiction. LLN holds as n → ∞, not finite.

Given p=0.3, n=50, sample mean = 0.46

LLN says about limit as n → ∞, not about n=50. So a deviation for finite n is not a contradiction.

σ √(p(1−p)/n) = √( /50 )      σ = 0.0648. Deviation: (0.46−0.3)/0.064 = 2.47 standard deviation, which is possible.

We will use the normal approximation here

P(X̄₅₀ ≥ 0.46) = p(z ≥ (0.46−0.3)/0.0648) = p(z ≥ 2.47)

Two tailed probablity = 0.0136 = 1.36%

It is about 1.4 times the probability by chance for deviation.

Such an event is rare but can happen. The LLN is a limit theorem, so one unusual sample at n=50 does not contradict the LLN

(2) The lifetime (in hours) of a certain type of electronic component has meanµ= 1200 hours and a standard deviationσ= 300 hours. Assume lifetimes of different components are independent and identically distributed.

A. [4pts] For = 25: [z̄]₂₅ = 1200, σ = 300/√25 = 60

(z̄₍₂₅₎ > 1300) = ( > 100/60)
                = ( > 1.67)
                = 1 − Φ(1.667)
                = 1 − 0.9525
                = 0.047

B. [4pts] For = 100: σ = 300/√100 = 30

(1150 < z̄₁₀₀ < 1250)
= ( − 1.67 < z < 1.67)
= 0.9525 − 0.0475 = 0.905

D. [4pts] Increasing from 25 to 100:

1 = 300 / √25 =

= 300 / √100 =

• decreases: 60 → 30 (50% reduction)
• Var() ∝ 1/
• CLT approximation improves (closer to normal)

The Central Limit Theorem approximation becomes more accurate for larger n, because the distribution of X̄ approaches normality more closely regardless of the population distribution’s shape. Less variability means more precise estimates of the population mean μ.

Q2 [17 pts] 1. In a baseband communication receiver, the input noise is modelled as an ideal white Gaussian noise process with two–sided PSD Sₓ(f) = N₀/2 for all f ∈ ℝ. The receiver front–end includes an ideal lowpass filter with bandwidth B Hz and unity gain in the passband.

A. [3pts] 0 = { 1,  |f| ≤
               { 0,  |f| >

|0|² = { 1,  |f| ≤
        { 0,  |f| >

Output PSD: 0 = { N₀/2,  |f| ≤
                 { 0,    |f| >

B. [3pts]

= ∫ 0
   −∞

= ∫ N₀/2
   −B   B

= N₀/2 × 2B

= N₀B

C. [3pts] Noise power ∝ B. Trade-off:

• Larger: more noise, worse SNR
• Decreasing B reduces output noise power
• Optimal: B = signal bandwidth (matched filtering)

Signal bandwidth: The filter must pass all important frequency components of the signal. If the signal has bandwidth Bs, we require B ≥ Bs to avoid signal distortion (e.g., intersymbol interference in digital systems, loss of fidelity in analogue).

Noise reduction: To maximise SNR, we want B as small as possible to minimise N₀B.

Trade-off: Choose B slightly larger than Bs to accommodate signal spectral roll-off and timing uncertainties, but not too large to avoid excess noise.

A. [6pts]

1. X₁(t): E[X₁(t)] = 0 (const),

Rₓ₁(t₁,t₂) = σ²e^{-|t₁−t₂|} = σ²e^{-|τ|}

Both conditions are satisfied. Therefore, it is WSS

2. X₂(t): E[X₂(t)] = A cos(2πf₀t) (time-varying) The mean is time dependent,

therefore

X₂(t) is NOT WSS

3. X₃(t): E[X₃(t)] = 0 (constant)

Rₓ₃(t₁,t₂) = σ²e^{−t₁}e^{−|t₂|} cannot be written as R(τ).

B. [2pts] Appropriate model for stationary background noise

1() is a good choice. Only the WSS process has time-invariant statistics. Zero mean, exponentially decaying correlation () = σ²e^{−|τ|} models realistic stationary noise

2 has a time-varying mean, not good for pure noise

3 is non-stationary, not appropriate