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Two probability distributions of the discrete random variable X and Y are given below. Prove that \[\mathbf{E}({{\mathbf{Y}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{E}\left( \mathbf{X} \right)\].
Two probability distributions of the discrete random variable X and Y are given below. Prove that \[\mathbf{E}({{\mathbf{Y}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{E}\left( \mathbf{X} \right)\].
CBSE | Class 12 | Exercise 1 | Maths | NCERT Exemplar | Probability
Two probability distributions of the discrete random variable X and Y are given below. Prove that \[\mathbf{E}({{\mathbf{Y}}^{\mathbf{2}}})\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{E}\left( \mathbf{X} \right)\].

The probability distribution of random variable X is

We know that,

\[E(X)=\sum\limits_{i=1}^{n}{{{P}_{i}}{{X}_{i}}}\]

\[E\left( X \right)\text{ }0.\text{ }1/5\text{ }+\text{ }1.\text{ }2/5\text{ }+\text{ }2.\text{ }1/5\text{ }+\text{ }3.\text{ }1/5\text{ }=\text{ }0\text{ }+\text{ }2/5\text{ }+\text{ }2/5\text{ }+\text{ }3/5\text{ }=\text{ }7/5\]

For the second probability distribution of random variable Y,

\[E({{Y}^{2}})\text{ }=\text{ }0.\text{ }1/5\text{ }+\text{ }1.\text{ }3/10\text{ }+\text{ }4.\text{ }2/5\text{ }+\text{ }9.\text{ }1/10\]

= \[0\text{ }+\text{ }3/10\text{ }+\text{ }8/5\text{ }+\text{ }9/10\text{ }=\text{ }28/10\text{ }=\text{ }14/5\]

Now, \[E({{Y}^{2}})\text{ }=\text{ }14/5\]and \[2\text{ }E\left( X \right)\text{ }=\text{ }2.7/5\text{ }=\text{ }14/5\]

Therefore, \[E({{Y}^{2}})\text{ }=\text{ }2E\left( X \right)\]

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