# Probability Homework Question 2

• MHB
• pinky14
In summary: The minimum will occur at the critical point $0$ or $1$, or at the boundary values $0$ and $1$. This means that $\alpha = \dfrac{\sigma_Y^2}{\sigma_X^2 + \sigma_Y^2}$ will minimize the variance of $W$.In summary, the conversation discusses finding the value of $\alpha$ in the interval [0, 1] that minimizes the variance of the random variable $W = \alpha X + (1 - \alpha)Y$, where $X$ and $Y$ are uncorrelated random variables with the same mean but different variances. The solution involves finding the critical point of a quadratic function and setting it equal to $0 pinky14 Suppose that X, Y are uncorrelated random variables which are each measurements of some unknown quantity$\mu$. Both random variables have$\mu_{X} = \mu_{Y} = \mu$, but$\sigma^2_{X} > \sigma^2_{Y}$. Determine the value of$\alpha$in [0, 1] which will minimize the variance of the random variable W =$\alpha X +(1 - \alpha)$Y. Note that E(W) =$\mu$for any$\alpha$so the minimal variance α gives the “best” linear combination of X, Y to use in estimating$\mu$. Hi pinky14, In the future, please show your thoughts or what you've tried. Show that$\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2\sigma_Y^2 + 2\alpha(1 - \alpha)\operatorname{Cov}(X,Y)$. Since$X$and$Y$are uncorrelated, what can you say about$\operatorname{Cov}(X,Y)$? Euge said: Hi pinky14, In the future, please show your thoughts or what you've tried. Show that$\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2\sigma_Y^2 + 2\alpha(1 - \alpha)\operatorname{Cov}(X,Y)$. Since$X$and$Y$are uncorrelated, what can you say about$\operatorname{Cov}(X,Y)$? So for 2 random variables to be "uncorrelated," their covariance has to be 0? (E(XY)-E(X)E(Y))? I wasn't really sure how to start with this problem. The problem says their means are equal but that the variance of X is larger than Y but I am not sure what to do with the equation for W that they give. When they say to minimize, do we take the derivative? I am pretty lost in my probability class. pinky14 said: So for 2 random variables to be "uncorrelated," their covariance has to be 0? That's correct. So then$\sigma_W^2 = \alpha^2 \sigma_X^2 + (1 - \alpha)^2 \sigma_Y^2$. You will need to find$\alpha\in [0,1]$that minimizes the quadratic function$f(\alpha) := \alpha^2 \sigma_X^2 + (1 - \alpha)^2 \sigma_Y^2\$.

## 1. What is the probability of getting a heads when flipping a fair coin?

The probability of getting a heads when flipping a fair coin is 0.5, or 50%. This is because there are only two possible outcomes (heads or tails) and they are equally likely to occur.

## 2. How do you calculate the probability of an event?

To calculate the probability of an event, you divide the number of favorable outcomes by the total number of possible outcomes. This gives you a decimal value, which can be converted to a percentage if desired.

## 3. What is the difference between theoretical and experimental probability?

Theoretical probability is based on mathematical calculations and assumes that all outcomes are equally likely to occur. Experimental probability is based on actual data from experiments or observations, and may differ from theoretical probability due to chance or other factors.

## 4. How does sample size affect the accuracy of probability calculations?

In general, larger sample sizes will result in more accurate probability calculations. This is because a larger sample size provides a more representative sample of the population and reduces the impact of random chance.

## 5. What is the difference between independent and dependent events?

Independent events are those in which the outcome of one event does not affect the outcome of another event. Dependent events are those in which the outcome of one event does affect the outcome of another event. This can impact the probability calculations for the events.

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