Respuesta :
Answer:
[tex]2\, c^{2}[/tex].
Step-by-step explanation:
The goal is to express [tex]\mathbb{E}(X^{2})[/tex] in terms of [tex]\mathbb{E}(X)[/tex] and [tex]{\rm Var}(X)[/tex].
Note that for a given random variable [tex]X[/tex], the following are equivalent:
- Variance [tex]{\rm Var}(X)[/tex] of this random variable;
- Expected value of the squared difference between [tex]X[/tex] and [tex]\mathbb{E}[X][/tex].
In other words:
[tex]\displaystyle {\rm Var}[X] = \mathbb{E}\left[[X - \mathbb{E}(X)]^{2}\right][/tex].
Expand the expression on the right side of this equation:
[tex]\begin{aligned} {\rm Var}[X] &= \mathbb{E}\left[\left[X - \mathbb{E}(X)\right]^{2}\right] \\ &= \mathbb{E}\left[X^{2} - X\, \mathbb{E}(X) + \left[\mathbb{E}(X)\right]^{2}\right]\end{aligned}[/tex].
Make use of the property that the expected value of a sum is equivalent to the sum of the expected value of the individual items. In other words, for any two random variables [tex]Y[/tex] and [tex]Z[/tex], [tex]\mathbb{E}(Y + Z) = \mathbb{E}(Y) + \mathbb{E}(Z)[/tex].
[tex]\begin{aligned} & \mathbb{E}\left[X^{2} - X\, \mathbb{E}(X) + \left[\mathbb{E}(X)\right]^{2}\right] \\ =\; & \mathbb{E}\left(X^{2}\right) - \mathbb{E}[2\, X\, \mathbb{E}(X)] + \mathbb{E}\left[\left[\mathbb{E}(X)\right]^{2}\right]\end{aligned}[/tex].
Similar to the "[tex]2[/tex]" in the expression, [tex]\mathbb{E}(X)[/tex] can be considered as a constant. Hence:
- [tex]\mathbb{E}[2\, X\, \mathbb{E}(X)] = (2)\, (\mathbb{E}[X])\, [\mathbb{E}(X)] = 2\, \left[\mathbb{E}(X)\right]^{2}[/tex].
- [tex]\mathbb{E}\left[\left[\mathbb{E}(X)\right]^{2}\right] = \left[\mathbb{E}(X)\right]^{2}[/tex].
[tex]\begin{aligned} & \mathbb{E}\left(X^{2}\right) - \mathbb{E}[2\, X\, \mathbb{E}(X)] + \mathbb{E}\left[\left[\mathbb{E}(X)\right]^{2}\right] \\ =\; & \mathbb{E}\left(X^{2}\right) - 2\, \left[\mathbb{E}(X)\right]^{2} + \left[\mathbb{E}(X)\right]^{2} \\ =\; & \mathbb{E}\left(X^{2}\right) - \left[\mathbb{E}(X)\right]^{2}\end{aligned}[/tex].
In other words:
[tex]\begin{aligned} {\rm Var}[X] &= \mathbb{E}\left[\left[X - \mathbb{E}(X)\right]^{2}\right] \\ &= \mathbb{E}\left(X^{2}\right) - \left[\mathbb{E}(X)\right]^{2}\end{aligned}[/tex].
Rearrange to obtain an expression for [tex]\mathbb{E}\left(X^{2}\right)[/tex]:
[tex]\begin{aligned}\mathbb{E}\left(X^{2}\right) &= {\rm Var}(X) + \left[\mathbb{E}(X)\right]^{2} \\ &= c^{2} + (c)^{2} \\ &= 2\, c^{2}\end{aligned}[/tex].
Final answer:
For an exponential random variable X with mean E[X]=c and variance Var[X]=c^2, the expected value of X^2 (E[X^2]) is found to be 2c^2 using the formula Var[X] = E[X^2] - (E[X])^2. Correct option is C 2c²
Explanation:
The question relates to a random variable X that follows an exponential distribution. Given that E[X]=c and Var[X]=c^2, we are asked to find E[X^2]. Recall that for an exponential distribution, the mean (E[X]) equals 1/lambda and the variance (Var[X]) equals 1/lambda^2, where lambda is the rate parameter. Therefore, with E[X]=c, we have lambda=1/c. By this relationship, Var[X], which is also 1/lambda^2, equals c^2 as given.
To find E[X^2], we use the fact that for any random variable, Var[X] = E[X^2] - (E[X])^2. Plugging in the values we have: c^2 = E[X^2] - c^2. Solving for E[X^2], we get E[X^2] = 2c^2, which matches option (c).
