We can show by the uniqueness theorem that the linear transformation, Y = aX + b, is also a normal random variable because the resultant probaility density fnction of Y equals: f(y) = (1/√(2πa^2σX^2)) * exp(-(y-aμX-b)^2/(2a^2σX^2)).
How to prove a normal random variableTo show that the linear transformation, Y = aX + b is a normal random variable, we need to demonstrate that it satisfies the properties of a normal distribution. This means that it should have a bell-shaped probability density function, mean, and variance.
We can prove that it meets the mean condition this way:
E(Y) = E(aX + b) = aE(X) + b = aμX + b
Next, we can prove that it meets the variance condition thus:
Var(Y) = Var(aX + b) = a^2Var(X) = a^2σX^2
Lastly, the probability density function is given as f(y) = (1/√(2πa^2σX^2)) * exp(-(y-aμX-b)^2/(2a^2σX^2)). This proves that the conditions for a normal random variable is met.
Learn more about a normal random variable here:
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^4√p7
in exponential form.
Answer:1111
Step-by-step explanation: