Uniform Distribution Check for Random Variable Transformations
The random variable X follows a discrete uniform distribution over the set $S_X = \{-10, -9, \dots, 9, 10\}$.
- The total number of possible values for X is $|S_X| = 10 - (-10) + 1 = 21$.
- The probability for each value is $P(X=k) = \frac{1}{21}$ for any integer $k$ such that $-10 \le k \le 10$.
A random variable is uniformly distributed if all its possible distinct outcomes have the same probability.
Analyzing Transformation $X^2$
Consider $Y_1 = X^2$. The possible values of $Y_1$ include $\{0, 1, 4, \dots, 100\}$.
- $P(Y_1 = 0) = P(X=0) = \frac{1}{21}$.
- $P(Y_1 = 1) = P(X^2=1) = P(X=1 \text{ or } X=-1) = P(X=1) + P(X=-1) = \frac{1}{21} + \frac{1}{21} = \frac{2}{21}$.
Since the probabilities $P(Y_1=0)$ and $P(Y_1=1)$ are different, $X^2$ is not uniformly distributed.
Analyzing Transformation $X^3$
Consider $Y_2 = X^3$. The function $f(x)=x^3$ is a one-to-one mapping from $S_X$ to the set of possible values for $Y_2$.
- The set of possible values is $S_{Y_2} = \{k^3 \mid k \in S_X\}$. The size of this set is also 21.
- For every distinct value $y$ in $S_{Y_2}$, there is exactly one $x$ in $S_X$ such that $x^3 = y$.
- Therefore, $P(Y_2 = y) = P(X = x) = \frac{1}{21}$.
Since all distinct outcomes have the same probability, $X^3$ is uniformly distributed.
Analyzing Transformation $(X-5)^2$
Consider $Y_3 = (X-5)^2$. Let $Z = X-5$. The possible values for $Z$ are $\{-15, -14, \dots, 5\}$.
- $P(Y_3 = 0) = P((X-5)^2=0) = P(X-5=0) = P(X=5) = \frac{1}{21}$.
- $P(Y_3 = 1) = P((X-5)^2=1) = P(X-5=1 \text{ or } X-5=-1) = P(X=6 \text{ or } X=4) = P(X=6) + P(X=4) = \frac{1}{21} + \frac{1}{21} = \frac{2}{21}$.
Since the probabilities $P(Y_3=0)$ and $P(Y_3=1)$ are different, $(X-5)^2$ is not uniformly distributed.
Analyzing Transformation $(X+10)^2$
Consider $Y_4 = (X+10)^2$. Let $W = X+10$. The possible values for $W$ are $\{0, 1, \dots, 20\}$.
The function $g(w) = w^2$ is one-to-one for $w \in \{0, 1, \dots, 20\}$ because all values are non-negative.
- The set of possible values is $S_{Y_4} = \{w^2 \mid w \in \{0, 1, \dots, 20\}\}$. The size of this set is 21.
- For every distinct value $y$ in $S_{Y_4}$, there is exactly one $w \in \{0, 1, \dots, 20\}$ such that $w^2 = y$.
- This unique $w$ corresponds to a unique $x = w-10$ in the original set $S_X$, since $w \in \{0, \dots, 20\}$ implies $x \in \{-10, \dots, 10\}$.
- Therefore, $P(Y_4 = y) = P(W=w) = P(X=x) = \frac{1}{21}$.
Since all distinct outcomes have the same probability, $(X+10)^2$ is uniformly distributed.
Uniform Distribution Conclusion
The random variables $X^3$ and $(X+10)^2$ are uniformly distributed.
Important Questions from Discrete Distributions
The value of a and b so that the following is probability mass function
X: 0 1 2 P(X = x): 3a 3b 4b with mean 1.1, is:
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be
S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and \(p\left( i \right) = \frac{1}{{561}}\left( {33 - i} \right)\)
. Let A denote the event that the even number of buffers are full. Then p(A) is :If X is a Poisson random variate with mean 3, then P(|X- 3| < 1) will be:
Let x ∼ N(μ, σ2) If μ2 = σ2, (μ > 0), then the value of P(X < -μ | X < μ) in terms of cumulative function N (0, 1) is:
Consider a binomial random variable X. If X1, X2,...Xn are independent and identically distributed samples from the distribution of X with sum \(Y = \mathop \sum \limits_{i = 1}^n {X_i}\) then the distribution of Y as n → ∞ can be approximated as.
