If sigma(t)^2 denotes the standard deviation in the quantity t^2, show that
sigma(t)^2=2tsigma(t)
I don't even know where to begin... help plzz
The formula cannot be correct as you have written it. sigma(t) and sigma(t^2) are constants that are characteristics of the entire distribution. Yet t is a variable. There would have to be a particular t for which the relationship holds. Is your t supposed to be the mean value?
To prove the given equation, we need to use the definition of standard deviation and some properties of variance. Let's start by breaking down the steps:
Step 1: Start with the definition of the variance.
The variance, Var(x), is defined as:
Var(x) = E[(x - E[x])^2]
where E[x] represents the expected value of x.
Step 2: Calculate the variance of the quantity t^2.
Using the equation from Step 1, we can rewrite it as:
Var(t^2) = E[(t^2 - E[t^2])^2]
Since E[t^2] = E[(t')^2], where t' represents a random variable with the same distribution as t, we can simplify it further as:
Var(t^2) = E[(t^2 - E[(t')^2])^2]
Step 3: Expand and simplify the equation.
Expanding the squared term inside the expectation, we get:
Var(t^2) = E[(t^2 - E[(t')^2]) * (t^2 - E[(t')^2])]
Expanding further, we have:
Var(t^2) = E[t^4 - 2t^2E[(t')^2] + (E[(t')^2])^2]
Step 4: Apply properties of expectation.
By linearity of expectation, we can split the expectation into individual terms:
Var(t^2) = E[t^4] - 2E[t^2E[(t')^2]] + E[(E[(t')^2])^2]
Since E[t^2E[(t')^2]] = E[t^2]E[(t')^2] (assuming t and t' are independent), we can rewrite it as:
Var(t^2) = E[t^4] - 2E[t^2]E[(t')^2] + (E[(t')^2])^2
Step 5: Calculate the variances.
The variance of t is given by:
Var(t) = E[(t - E[t])^2]
The variance of t' (with the same distribution as t) can be written as:
Var(t') = E[(t' - E[t'])^2]
Since E[t'] = E[t], we can simplify it further as:
Var(t') = E[(t' - E[t])^2]
= E[(t - E[t])^2]
= Var(t)
Therefore, we can replace E[(t')^2] with E[t^2] in the equation above:
Var(t^2) = E[t^4] - 2E[t^2]E[t^2] + (E[t^2])^2
Step 6: Simplify the equation.
By combining like terms, we get:
Var(t^2) = E[t^4] - E[t^2]^2
Step 7: Use the definition of variance.
The standard deviation is the square root of the variance:
sigma(t)^2 = Var(t) = E[t^2] - E[t]^2
Step 8: Substitute the equation from Step 6.
sigma(t)^2 = E[t^2] - E[t]^2
= Var(t^2)
Now we have proven that sigma(t)^2 is equal to Var(t^2). Hence:
sigma(t)^2 = Var(t^2) = E[t^4] - E[t^2]^2
Step 9: Simplify the equation further.
We can rewrite t^4 as (t^2)^2:
sigma(t)^2 = E[(t^2)^2] - E[t^2]^2
Finally, comparing this to the given equation sigma(t)^2 = 2t * sigma(t), we can see that E[(t^2)^2] is equivalent to 2t * E[t^2].
Thus, sigma(t)^2 = 2tsigma(t).