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Data Interview Question

Random Seed Utilization

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Solution & Explanation

Understanding the Problem:

We have two functions:

  • Function X: Generates a random integer between a minimum value, NN, and a maximum value, MM.
  • Function Y: Uses the output of Function X as its maximum value, with the same minimum value NN.

Distribution of Samples from Function Y:

  1. Function X Distribution:

    • XX is a discrete uniform random variable over [N,M][N, M].
    • Probability Mass Function (PMF) of XX: pX(x)=1MN+1forx[N,M]p_X(x) = \frac{1}{M - N + 1}\quad \text{for}\quad x \in [N, M]
  2. Function Y Distribution Given X=xX = x:

    • YY is a discrete uniform random variable over [N,x][N, x].
    • Conditional PMF of YY given X=xX = x: pYX(yx)=1xN+1fory[N,x]p_{Y|X}(y|x) = \frac{1}{x - N + 1}\quad \text{for}\quad y \in [N, x]
  3. Joint Distribution of XX and YY:

    • pX,Y(x,y)=pX(x)pYX(yx)p_{X,Y}(x,y) = p_X(x) \cdot p_{Y|X}(y|x)
    • pX,Y(x,y)=1MN+11xN+1p_{X,Y}(x,y) = \frac{1}{M - N + 1} \cdot \frac{1}{x - N + 1}
  4. Marginal Distribution of YY:

    • To find the marginal distribution pY(y)p_Y(y), we integrate over all possible values of XX: pY(y)=x=yMpX,Y(x,y)=x=yM1(MN+1)(xN+1)p_Y(y) = \sum_{x=y}^{M} p_{X,Y}(x,y) = \sum_{x=y}^{M} \frac{1}{(M-N+1)(x-N+1)}
    • This results in a non-uniform distribution for YY, as the probability of YY decreases as yy approaches MM.

Expected Value of Samples from Function Y:

  1. Using the Law of Iterated Expectation:

    • E[Y]=E[E[YX]]E[Y] = E[E[Y|X]]
  2. Expected Value of YY Given X=xX = x:

    • E[YX=x]=N+x2E[Y|X=x] = \frac{N + x}{2}
  3. Overall Expected Value of YY:

    • E[Y]=E[N+X2]=12(N+E[X])E[Y] = E\left[\frac{N + X}{2}\right] = \frac{1}{2} \left(N + E[X]\right)
    • E[X]=M+N2E[X] = \frac{M + N}{2} (since XX is uniform over [N,M][N, M])
    • E[Y]=12(N+M+N2)=3N+M4E[Y] = \frac{1}{2} \left(N + \frac{M + N}{2}\right) = \frac{3N + M}{4}

Conclusion:

  • Distribution of YY: Non-uniform, with probabilities decreasing as yy approaches MM.
  • Expected Value of YY: 3N+M4\frac{3N + M}{4}, which reflects the influence of both the uniform distribution of XX and the conditional constraints on YY.

The problem illustrates how conditional distributions and expectations can be derived from basic principles of probability and statistics, demonstrating the non-uniform nature of YY despite XX being uniformly distributed.