// Numbas version: finer_feedback_settings {"name": "PA3 - Expected Value & Variance", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "PA3 - Expected Value & Variance", "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "
The probability density function of a random variable is:
\n$f_Y(y) = \\begin{cases} \\var{function} & 0 \\le y \\le 1, \\\\ 0 & \\text{otherwise} \\end{cases}$
", "advice": "{advice}
\nUse this link to find some resources which will help you revise this topic.
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", "templateType": "mathematical expression", "can_override": false}, "prob0": {"name": "prob0", "group": "Function0", "definition": "0.25", "description": "P(y<0.5)
", "templateType": "anything", "can_override": false}, "expected0": {"name": "expected0", "group": "Function0", "definition": "expression(safe(\"2/3\"))", "description": "Expected value
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", "templateType": "mathematical expression", "can_override": false}, "advice0": {"name": "advice0", "group": "Function0", "definition": "\"a) To calculate $P(y<0.5)$ we must solve $\\\\int_0^{0.5} 2y dy.$
\\nSo,
\\n\\\\begin{align}\\\\int_0^{0.5} 2y dy &= \\\\left[y^2\\\\right]_0^{0.5} \\\\\\\\
&= (0.5)^{2} - (0)^{2} \\\\\\\\
&= 0.25 \\\\end{align}
Hence, $P(y<0.5) = 0.25.$
\\nb) The formula for expected value is:
\\n$E[Y] = \\\\int_{-\\\\infty}^{\\\\infty} y.f(y) dy.$
So,
\\\\begin{align}
E[Y] &= \\\\int_0^{1}y.2y dy \\\\\\\\
&= \\\\int_0^{1} 2y^2 dy \\\\\\\\
&= \\\\left[\\\\frac{2}{3}y^3\\\\right]_0^{1} \\\\\\\\
&= \\\\frac{2}{3}\\\\times1^3 - \\\\frac{2}{3} \\\\times 0^3 \\\\\\\\
&= \\\\frac{2}{3}.
\\\\end{align}
Hence, $E[Y] = \\\\frac{2}{3}$.
c) Our formula for $E[Y^2]$ is:
$E\\\\left[Y^2\\\\right] = \\\\int_{-\\\\infty}^{\\\\infty} y^2.f(y) dy.$
So,
\\n\\\\begin{align}
E\\\\left[Y^2\\\\right] &= \\\\int_{0}^{1} y^2.2y dy \\\\\\\\
&= \\\\int_{0}^{1} 2y^3 dy \\\\\\\\
&= \\\\left[\\\\frac{2}{4}y^4\\\\right]_0^{1} \\\\\\\\
&= \\\\frac{2}{4}\\\\times1^4 - \\\\frac{2}{4} \\\\times 0^4 \\\\\\\\
&= \\\\frac{2}{4} \\\\\\\\
&= 0.5.
\\\\end{align}
Hence, $E\\\\left[Y^2\\\\right] = 0.5$.
\\nd) Now we can use everything we\\'ve calculated so far to find the variance,
\\n\\\\begin{align}
E\\\\left[Y^2\\\\right] - E[Y]^2 &= 0.5 - \\\\left(\\\\frac{2}{3}\\\\right)^2 \\\\\\\\
&= 0.5 - \\\\frac{4}{9} \\\\\\\\
&= \\\\frac{1}{18}.
\\\\end{align}
Hence, $E\\\\left[Y^2\\\\right] - E[Y]^2 = \\\\frac{1}{18}$.
\"", "description": "Advice
", "templateType": "long string", "can_override": false}, "advice": {"name": "advice", "group": "Ungrouped variables", "definition": "if(random=0,advice0,if(random=1,advice1,0))", "description": "Advice for selected function
", "templateType": "anything", "can_override": false}, "advice1": {"name": "advice1", "group": "Function1", "definition": "\"a) To calculate $P(y<0.5)$ we must solve $\\\\int_0^{0.5}3y^2 dy.$
\\nSo,
\\n\\\\begin{align}\\\\int_0^{0.5}3y^2 dy &= \\\\left[\\\\frac{3}{3}y^3\\\\right]_0^{0.5} \\\\\\\\
&= (0.5)^3 - (0)^3 \\\\\\\\
&= 0.125 . \\\\end{align}
Hence, $P(y<0.5) = 0.125.$
\\nb) The formula for expected value is:
\\n$E[Y] = \\\\int_{-\\\\infty}^{\\\\infty} y.f(y) dy.$
So,
\\\\begin{align}
E[Y] &= \\\\int_0^{1}y.3y^2 dy \\\\\\\\
&= \\\\int_0^{1}3y^3 dy \\\\\\\\
&= \\\\left[\\\\frac{3}{4}y^4\\\\right]_0^{1} \\\\\\\\
&= \\\\frac{3}{4}\\\\times1^4 - \\\\frac{3}{4} \\\\times 0^4 \\\\\\\\
&= \\\\frac{3}{4} \\\\\\\\
&= 0.75.
\\\\end{align}
Hence, $E[Y] = 0.75$.
c) Our formula for $E[Y^2]$ is:
$E\\\\left[Y^2\\\\right] = \\\\int_{-\\\\infty}^{\\\\infty} y^2.f(y) dy.$
So,
\\n\\\\begin{align}
E\\\\left[Y^2\\\\right] &= \\\\int_{0}^{1} y^2.3y^2 dy \\\\\\\\
&= \\\\int_{0}^{1} 3y^4 dy \\\\\\\\
&= \\\\left[\\\\frac{3}{5}y^5\\\\right]_0^{1} \\\\\\\\
&= \\\\frac{3}{5}\\\\times1^5 - \\\\frac{1}{5} \\\\times 0^5 \\\\\\\\
&= \\\\frac{3}{5} \\\\\\\\
& = 0.6.
\\\\end{align}
Hence, $E\\\\left[Y^2\\\\right] = 0.6 $.
\\nd) Now we can use everything we\\'ve calculated so far to find the variance,
\\n\\\\begin{align}
E\\\\left[Y^2\\\\right] - E[Y]^2 &= 0.6 - 0.75^2 \\\\\\\\
&= 0.0375. \\\\\\\\
\\\\end{align}
Hence, $E\\\\left[Y^2\\\\right] - E[Y]^2 = 0.0375$.
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