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  • . Two probability distributions of the discrete random variable X and Y are given below.X 0 P(X) 1/5 Prove that E(Y)2=2E(X).

Two probability distributions of the discrete random variable X and Y are given below.

\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \\ \hline \mathrm{Y} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{y}) & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \frac{1}{10} \\ \hline \end{array}

Prove that E(Y^2)=2E(X).

Answers (1)

Solution

To prove that- E(Y)^2=2E(X)

Taking LHS of equation (i), we have:
E(Y^{2})=Y^{2} P(Y)
\\=0 \times \frac{1}{5}+1 \times \frac{3}{10}+4 \times \frac{2}{5}+9 \times \frac{1}{10} \\=\frac{3}{10}+\frac{8}{5}+\frac{9}{10}=\frac{28}{10}=\frac{14}{5} \\\Rightarrow \mathrm{E}(\mathrm{Y}^2)=\frac{14}{5}
Taking RHS of equation (i) we get:
E(X)=X P(X)

\\ =0 \times \frac{1}{5}+1 \times \frac{2}{5}+3 \times \frac{1}{5}=\frac{7}{5} \\2 \mathrm{E}(\mathrm{X})=2\left(\frac{7}{5}\right) \\=\frac{14}{5}
Thus, from equations (ii) and (iii), we get:
E\left(Y^{2}\right)=2 E(X)
Hence proved.

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