If X~N(3, 4), Y~N(-2, 6), Z~N(1, 3), and X and Y and Z are independent, what are the distribution, expectation and variance of S = 3X-2Y+Z? Notation: N(3, 4) is a Normal RV with mean 3 and variance 4.

S~N(13, 63)

Distribution = normal

Expected value = mean = 14

Variance = 63

Step-by-step explanation:

The theory of Sums of Independent Normal Random Variables explains that if X₁, X₂, X₃..., Xₙ represent normal random variables that are all mutually independent and have means μ₁, μ₂, μ₃, μₙ and variances, σ₁², σ₂², σ₃², σₙ² that are linearly combined in the form

Y = Σ kᵢXᵢ (summing from i = 1, to i = n)

Gives a normal random variable that Has a Mean of

μ = Σ kᵢμᵢ (summing from i = 1, to i = n)

And a variance

σ² = Σ kᵢ²σᵢ² (summing from i = 1, to i = n)

So, X~N(3, 4), Y~N(-2, 6), Z~N(1, 3) are linearly combined as

S = 3X-2Y+Z

The new mean of the combination

μ = Σ kᵢμᵢ (summing from i = 1, to i = n)

μ = (3)(3) + (-2)(-2) + (1)(1) = 9 + 4 + 1 = 14

And the variance is

σ² = Σ kᵢ²σᵢ² (summing from i = 1, to i = n)

σ² = (3²)(4) + (-2)²(6) + (1²)(3) = 36 + 24 + 3 = 63

S~N(13, 63)

Distribution = normal

Expected value = mean = 14

Variance = 63

Hope this Helps!!!

your answer would be 12