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Let U and V be two independent zero mean Gaussian random variables of variances 1/4 and 1/9 respectively. The probability P(3V ≥ 2U) is
- a)4/9
- b)2/3
- c)1/2
- d)5/9
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Let U and V be two independent zero mean Gaussian random variables of ...
Let X = 3V - 2U. Then X is also a Gaussian random variable with mean 0 and variance:
Var(X) = 9Var(V) + 4Var(U) = 9/4 + 4/9 = 97/36.
To find P(3V < 2u),="" we="" can="" rewrite="" this="" />
P(3V - 2U < />
Dividing both sides by sqrt(Var(X)), we get:
P((3V - 2U)/sqrt(97/36) < />
Let Z = (3V - 2U)/sqrt(97/36). Then Z is a standard normal random variable (i.e., with mean 0 and variance 1). So we want to find:
P(Z < />
This probability is just 0.5, since the standard normal distribution is symmetric about 0. Therefore:
P(3V < 2u)="P(Z" />< 0)="0.5." 0)="" />
Var(X) = 9Var(V) + 4Var(U) = 9/4 + 4/9 = 97/36.
To find P(3V < 2u),="" we="" can="" rewrite="" this="" />
P(3V - 2U < />
Dividing both sides by sqrt(Var(X)), we get:
P((3V - 2U)/sqrt(97/36) < />
Let Z = (3V - 2U)/sqrt(97/36). Then Z is a standard normal random variable (i.e., with mean 0 and variance 1). So we want to find:
P(Z < />
This probability is just 0.5, since the standard normal distribution is symmetric about 0. Therefore:
P(3V < 2u)="P(Z" />< 0)="0.5." 0)="" />
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Community Answer
Let U and V be two independent zero mean Gaussian random variables of ...
U and V are two independent zero mean and Gaussian.
let z = 3V - 2U
U and V are Gaussian then their linear transformation Z is also Gaussian
than E[z] = E[3V - 2U]=3E[V] - 2E[U]=0
if z is gaussian and zero mean then its probability for greater than zero is 0.5
Option C is the correct answer.
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Let U and V be two independent zero mean Gaussian random variables of variances 1/4 and 1/9 respectively. The probability P(3V ≥ 2U) isa)4/9b)2/3c)1/2d)5/9Correct answer is option 'C'. Can you explain this answer?
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