An Expected Value

Tamara can choose $425$ numbers from $1$ to $425$ .

Let $x_i$ denote the number selected such that $x_1=1, \ldots , x_{425}=425$ .

The expected value of $y$ is the average of all the expected values of $y_i$ for every corresponding $x_i$ .

We calculate that $E (y_1)=1, E(y_2)=1.5, E (y_3)=2,...., E (y_{425})=213$

And,

$E (y)=\frac {1}{425} \displaystyle \sum_{i=1}^{425} E (y_i)$

$E (y)=\frac {1}{425} \cdot 425 \cdot (\frac {1+213}{2})$ [Note that the summation yields this expression since first term is $1$ and the last $425$ with common difference $0.5$ ].

Hence, $E (y) =\boxed { 107}$

For a given $x$ , $\displaystyle E(y) = \frac{\sum_{i=1}^x i}{x} = \frac{\frac{x(x+1)}{2}}{x} = \frac{x+1}{2}$ .

Since $x$ is uniform randomly selected from integers in $[1, 425]$ , the probability that $x$ takes on any unique integer value in the specified interval is $\frac{1}{425}$ .

Consequently we have $E(y) = \displaystyle \frac{\sum_{n=1}^{425} \frac{n+1}{2}}{425} = \frac{\sum_{n=1}^{425} n+1}{850}$ $= \frac{\frac{425 \times 426}{2} + 425}{850} = \frac{\frac{1 \times 426}{2} + 1}{2} = \frac{214}{2} = \boxed{107}$ .

The expected value of y for a given x is (x+1)/2. Sum this for 1<=x<=425, and divide by 425 to get the overall average. Here are some terms: 1/425*(1+1.5+2+... 426/2). There are 425 terms, each term averaging 214/2 (pair up the first and last terms, 2nd & 424th etc), so the total is 214/2 = 107.

When Tamara chooses the number $x,$ the expected value of $y$ is $\frac{x+1}{2}.$ Thus, the expected value of $y$ is the expected value of $\frac{x+1}{2}$ as $x$ ranges uniformly from $1$ to $425.$ By linearity of expectation, $E[y] = E[\frac{x+1}{2}] = \frac{E[x] + 1}{2}.$ Since $E[x] = \frac{ 425+1}{2} = 213,$ hence $E[y] = \frac{213+1}{2} = 107.$

By Iterated Expectation theorem:

$E[Y] = E[E[Y|X]]$

Considering this problem,

$E[Y|X=x] = \frac{1+x}{2}$

(The above is true because $Y$ is Uniform Distribution which takes values between $1$ and $x$ .

In general the expected value of random variable $X$ having Uniform Distribution is $\frac{a+b}{2}$ where $X$ takes values from $a$ to $b$ .)

Therefore, in general case where $X$ can take any value,

$E[Y|X] = \frac{1+X}{2}$

Now,

$E[Y] = E[E[Y|X]]$

$= E[\frac{1+X}{2}] = \frac{1}{2} + \frac{1}{2}E[X]$

(by Linearity of Expectation )

$= 0.5 + 0.5 * \frac{1+425}{2} = 107$

$\frac{a+3}{4}$ where a = 425

Generally, when L is number of chooses (how deep): $\frac{a+2^L-1}{2^L}$

Firstly, to find the expected value of x, I multiplied the probability of each number being picked by the value of that number. Since each number has the probability of being picked P(x) = 1/425, and retains its own integer value, the expected value of x is (1/425) (1 + 2 + 3 ... + 424 + 425) = (425 426)/(425*2) = 213.

Essentially doing the same process with y except now P(y) = 1/213, we get the expected value of y as (1/213) * (1 + 2 + 3 + .... + 212 + 213) = (213 214)/(213 2) = 107.

Therefore the EV of y is 107.

First of all we need to find $E(x)$ . $E(x)=(1+2+3+............+425)/425 =(425*426)/(2*425) =213$ . Now, $E(y)=(1+2+3+..........+213)/213 =(213*214)/(2*213) =107$ . Hence the expected value of $y= \underline{107}$ .

If a random variable is uniformly distributed in $\{1,2,\dots,n\}$ , then its expected value is given by $\sum_{k=1}^{n}\frac{k}{n} = \frac{n+1}{2}$ . Then, we use that to write $\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y|X]] = \mathbb{E}\left[\frac{1+X}{2}\right] = \frac{1+\frac{1+425}{2}}{2} = 107.$

Just a tease.

To find the expected value of x, you must find the median value of x: \frac {415 + 1}{2} = 213 Now, we do the same to find the expected value of y: \frac {213 + 1}{2} = 107 Therefore, the expected value for y is 107.

The expected value of $x = \frac{1+425}{2} = 213$ then the expected value of $y = \frac{1+213}{2} = 107$

Here we have to understand the language of problem statement. Choosing a random number from 1 to 425, "UNIFORMLY". Here uniformly means at constant differences. So the series formed will be A.P

Hence applying (x)=(d/2) (2 a+(n-1)*d) where S(x) is the sum of series up-to n d is the constant difference.

Here, S(x)=(1/2) (2 1+(425-1)*1)=213

Now, Once again we have to repeat the process, because selection of number choosing "UNIFORMLY" at random is repeated for new value of n(previous result). S'(x)=(1/2) (2 1+(213-1)*1)=107

here, we have answer as 107.

* The same problem can be for other than uniformly, it can be G.P